Embed Size (px)
Citation preview
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
John P Shen amp Zhiyi YuSeptember 26 2016
9262016 (copyJP Shen) 18-600 Lecture 8 1
18-600 Foundations of Computer Systems
CS AAP
CS APP
18-600
Required Reading Assignmentbull Chapter 4 of CSAPP (3rd edition) by Randy Bryant amp Dave OrsquoHallaron
Recommended Reference Chapters 1 and 2 of Shen and Lipasti (SnL)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 2
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ComputerEngineering
Application Softwarebull Program developmentbull Program compilation
Hardware Technologybull Program executionbull Computer performance
Computer Instruction Set Architecture
9262016 (copyJP Shen) 18-600 Lecture 8 3
Instruction Set Architecture (ISA) SPECIFICATION
IMPLEMENTATION
SoftwareEngineering
Proc
esso
rDe
sign
Prog
ram
Deve
lopm
ent
CS APP
CS AAP
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Set Architecture Assembly Language View
bull Processor state (architecture state)bull Registers Memory hellip
bull Instructions (specify operations operands)bull addq pushq ret hellipbull How instructions are encoded as bytes
Layer of Abstractionbull Above how to program machine
bull Processor executes instructions in a sequencebull Below what needs to be implemented
bull Use variety of uarch techniques to make it run fastbull Eg execute multiple instructions simultaneously
ISA
Compiler OS
CPUDesign
CircuitDesign
ChipLayout
ApplicationProgram
9262016 (copyJP Shen) 18-600 Lecture 8 4
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CISC Instruction Sets Complex Instruction Set Computer IA32 is example
Stack-oriented instruction set Use stack to pass arguments save program counter Explicit push and pop instructions
Arithmetic instructions can access memory addq rax 12(rbxrcx8)
requires memory read and write Complex address calculation
Condition codes Set as side effect of arithmetic and logical instructions
Philosophy Add instructions to perform ldquotypicalrdquo programming tasks
9262016 (copyJP Shen) 18-600 Lecture 8 5
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
RISC Instruction Sets Reduced Instruction Set Computer Internal project at IBM later popularized by Hennessy (Stanford) and Patterson (Berkeley)
Fewer simpler instructions Might take more to get given task done Can execute them with small and fast hardware
Register-oriented instruction set Many more (typically 32) registers Use for arguments return pointer temporaries
Only load and store instructions can access memory Similar to Y86-64 mrmovq and rmmovq
No Condition codes Test instructions return 01 in register
9262016 (copyJP Shen) 18-600 Lecture 8 6
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Quantum Physics
Transistors amp Devices
Logic Gates amp Memory
Von Neumann Machine
x86 Machine Primitives
Visual C++
Firefox MS Excel
Windows
Computer Dynamic-Static Interface
9262016 (copyJP Shen) 18-600 Lecture 8 7
DynamicStatic Interface (DSI)=(ISA)
ldquosta
ticrdquo
ldquodyn
amic
rdquo
Architectural State
Microarchitecture State
MACHINE
PROGRAM
Exposed to SW
Hidden in HW
Buffering needed between arch and uarch statesbull Allow uarch state to deviate from arch statebull Able to undo speculative uarch state if needed
Architectural state requirementsbull Support sequential instruction execution semanticsbull Support precise servicing of exceptions amp interrupts
ARCHITECTURE
DSI = ISA = a contract between the program and the machine
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CISC vs RISCOriginal Debate
Strong opinions CISC proponents---easy for compiler fewer code bytes RISC proponents---better for optimizing compilers can make machine run fast with
simple chip design
Current Status For desktop processors choice of ISA not a technical issue
With enough hardware can make anything run fast Code compatibility more important
x86-64 adopted many RISC features More registers use them for argument passing
For embedded processors RISC makes sense Smaller cheaper less power Most mobile phones use ARM processors
9262016 (copyJP Shen) 18-600 Lecture 8 8
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
MIPS Registers$0$1$2$3$4$5$6$7$8$9$10$11$12$13$14$15
$0$at$v0$v1$a0$a1$a2$a3$t0$t1$t2$t3$t4$t5$t6$t7
Constant 0Reserved Temp
Return Values
Procedure arguments
Caller SaveTemporariesMay be overwritten bycalled procedures
$16$17$18$19$20$21$22$23$24$25$26$27$28$29$30$31
$s0$s1$s2$s3$s4$s5$s6$s7$t8$t9$k0$k1$gp$sp$s8$ra
Reserved forOperating Sys
Caller Save Temp
Global Pointer
Callee SaveTemporariesMay not beoverwritten bycalled procedures
Stack PointerCallee Save TempReturn Address
9262016 (copyJP Shen) 18-600 Lecture 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
MIPS Instruction Examples
Op Ra Rb Offset
Op Ra Rb Rd Fn00000R-R
Op Ra Rb ImmediateR-I
LoadStore
addu $3$2$1 Register add $3 = $2+$1
addu $3$2 3145 Immediate add $3 = $2+3145
sll $3$22 Shift left $3 = $2 ltlt 2
lw $316($2) Load Word $3 = M[$2+16]
sw $316($2) Store Word M[$2+16] = $3
Op Ra Rb OffsetBranch
beq $3$2dest Branch when $3 = $2
9262016 (copyJP Shen) 18-600 Lecture 8 10
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Evolutions of MIPS and ARM ISAs
9262016 (copyJP Shen) 18-600 Lecture 8 11
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ISA Design Summary
How Important is ISA Design Less now than before
With enough hardware can make almost anything go fast Installed based of legacy code and SW development tools dominate
Y86-64 Instruction Set Architecture Similar state and instructions as x86-64 Simpler encodings Somewhere between CISC and RISC
9262016 (copyJP Shen) 18-600 Lecture 8 12
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Processor Statebull Program Registers
bull 15 registers (omit r15) Each 64 bits
bull Condition Codesbull Single-bit flags set by arithmetic or logical instructions
bull ZF Zero SFNegative OF Overflow
bull Program Counterbull Indicates address of next instruction
bull Program Statusbull Indicates either normal operation or some error condition
bull Memorybull Byte-addressable storage arraybull Words stored in little-endian byte order
ZFSFOF
RF Program registers
CC Condition
codes
PC
DMEM Memory
Stat Program status
r8
r9
r10
r11
r12
r13
r14
rax
rcx
rdx
rbx
rsp
rbp
rsi
rdi
9262016 (copyJP Shen) 18-600 Lecture 8 13
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Set
18-600 Lecture 89262016 (copyJP Shen) 14
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Formats
Formatbull 1ndash10 bytes of information read from memory
bull Can determine instruction length from first bytebull Fewer instruction types and simpler encoding than x86-64
bull Each instruction accesses and modifies some part(s) of the program state
9262016 (copyJP Shen) 18-600 Lecture 8 15
18-600 Lecture 89262016 (copyJP Shen) 16
0 1 2 3 4 5 6 7 8 9
V
D
D
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB
rmmovq rA D(rB) 4 0 rA rB
mrmovq D(rB) rA 5 0 rA rB
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
rrmovq 2 0
cmovle 2 1
cmovl 2 2
cmove 2 3
cmovne 2 4
cmovge 2 5
cmovg 2 6
Y86-64 Instructions (moves)
18-600 Lecture 89262016 (copyJP Shen) 17
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
addq 6 0
subq 6 1
andq 6 2
xorq 6 3
Y86-64 Instructions (ALUs)
18-600 Lecture 89262016 (copyJP Shen) 18
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9jmp 7 0
jle 7 1
jl 7 2
je 7 3
jne 7 4
jge 7 5
jg 7 6
Y86-64 Instructions (branches)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Encoding Register Specifiers
Each register has 4-bit ID
bull Same encoding as in x86-64 Register ID 15 (0xF) indicates ldquono registerrdquo
bull Will use this in our hardware design in multiple places
rax
rcx
rdx
rbx
0
1
2
3
rsp
rbp
rsi
rdi
4
5
6
7
r8
r9
r10
r11
8
9
A
B
r12
r13
r14
No Register
C
D
E
F
9262016 (copyJP Shen) 18-600 Lecture 8 19
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Format ExampleAddition Instruction
Add value in register rA to that in register rB Store result in register rB Note that Y86-64 only allows addition to be applied to register data
Set condition codes based on result eg addq raxrsi Encoding 60 06
Two-byte encoding First indicates instruction type Second gives source and destination registers
addq rA rB 6 0 rA rB
Encoded Representation
Generic Form
9262016 (copyJP Shen) 18-600 Lecture 8 20
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Arithmetic and Logical Operations
Refer to generically as ldquoOPqrdquo Encodings differ only by ldquofunction coderdquo
Low-order 4 bits in first instruction byte
Set condition codes as side effect
addq rA rB 6 0 rA rB
subq rA rB 6 1 rA rB
andq rA rB 6 2 rA rB
xorq rA rB 6 3 rA rB
Add
Subtract (rA from rB)
And
Exclusive-Or
Instruction Code Function Code
9262016 (copyJP Shen) 18-600 Lecture 8 21
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Operations
Like the x86-64 movq instruction Simpler format for memory addresses Give different names to keep them distinct
rrmovq rA rB 2 0
Register Register
Immediate Register
irmovq V rB F rB3 0 V
Register Memory
rmmovq rA D(rB) 4 0 rA rB D
Memory Register
mrmovq D(rB) rA 5 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 22
rA rB
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 2
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ComputerEngineering
Application Softwarebull Program developmentbull Program compilation
Hardware Technologybull Program executionbull Computer performance
Computer Instruction Set Architecture
9262016 (copyJP Shen) 18-600 Lecture 8 3
Instruction Set Architecture (ISA) SPECIFICATION
IMPLEMENTATION
SoftwareEngineering
Proc
esso
rDe
sign
Prog
ram
Deve
lopm
ent
CS APP
CS AAP
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Set Architecture Assembly Language View
bull Processor state (architecture state)bull Registers Memory hellip
bull Instructions (specify operations operands)bull addq pushq ret hellipbull How instructions are encoded as bytes
Layer of Abstractionbull Above how to program machine
bull Processor executes instructions in a sequencebull Below what needs to be implemented
bull Use variety of uarch techniques to make it run fastbull Eg execute multiple instructions simultaneously
ISA
Compiler OS
CPUDesign
CircuitDesign
ChipLayout
ApplicationProgram
9262016 (copyJP Shen) 18-600 Lecture 8 4
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CISC Instruction Sets Complex Instruction Set Computer IA32 is example
Stack-oriented instruction set Use stack to pass arguments save program counter Explicit push and pop instructions
Arithmetic instructions can access memory addq rax 12(rbxrcx8)
requires memory read and write Complex address calculation
Condition codes Set as side effect of arithmetic and logical instructions
Philosophy Add instructions to perform ldquotypicalrdquo programming tasks
9262016 (copyJP Shen) 18-600 Lecture 8 5
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
RISC Instruction Sets Reduced Instruction Set Computer Internal project at IBM later popularized by Hennessy (Stanford) and Patterson (Berkeley)
Fewer simpler instructions Might take more to get given task done Can execute them with small and fast hardware
Register-oriented instruction set Many more (typically 32) registers Use for arguments return pointer temporaries
Only load and store instructions can access memory Similar to Y86-64 mrmovq and rmmovq
No Condition codes Test instructions return 01 in register
9262016 (copyJP Shen) 18-600 Lecture 8 6
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Quantum Physics
Transistors amp Devices
Logic Gates amp Memory
Von Neumann Machine
x86 Machine Primitives
Visual C++
Firefox MS Excel
Windows
Computer Dynamic-Static Interface
9262016 (copyJP Shen) 18-600 Lecture 8 7
DynamicStatic Interface (DSI)=(ISA)
ldquosta
ticrdquo
ldquodyn
amic
rdquo
Architectural State
Microarchitecture State
MACHINE
PROGRAM
Exposed to SW
Hidden in HW
Buffering needed between arch and uarch statesbull Allow uarch state to deviate from arch statebull Able to undo speculative uarch state if needed
Architectural state requirementsbull Support sequential instruction execution semanticsbull Support precise servicing of exceptions amp interrupts
ARCHITECTURE
DSI = ISA = a contract between the program and the machine
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CISC vs RISCOriginal Debate
Strong opinions CISC proponents---easy for compiler fewer code bytes RISC proponents---better for optimizing compilers can make machine run fast with
simple chip design
Current Status For desktop processors choice of ISA not a technical issue
With enough hardware can make anything run fast Code compatibility more important
x86-64 adopted many RISC features More registers use them for argument passing
For embedded processors RISC makes sense Smaller cheaper less power Most mobile phones use ARM processors
9262016 (copyJP Shen) 18-600 Lecture 8 8
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
MIPS Registers$0$1$2$3$4$5$6$7$8$9$10$11$12$13$14$15
$0$at$v0$v1$a0$a1$a2$a3$t0$t1$t2$t3$t4$t5$t6$t7
Constant 0Reserved Temp
Return Values
Procedure arguments
Caller SaveTemporariesMay be overwritten bycalled procedures
$16$17$18$19$20$21$22$23$24$25$26$27$28$29$30$31
$s0$s1$s2$s3$s4$s5$s6$s7$t8$t9$k0$k1$gp$sp$s8$ra
Reserved forOperating Sys
Caller Save Temp
Global Pointer
Callee SaveTemporariesMay not beoverwritten bycalled procedures
Stack PointerCallee Save TempReturn Address
9262016 (copyJP Shen) 18-600 Lecture 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
MIPS Instruction Examples
Op Ra Rb Offset
Op Ra Rb Rd Fn00000R-R
Op Ra Rb ImmediateR-I
LoadStore
addu $3$2$1 Register add $3 = $2+$1
addu $3$2 3145 Immediate add $3 = $2+3145
sll $3$22 Shift left $3 = $2 ltlt 2
lw $316($2) Load Word $3 = M[$2+16]
sw $316($2) Store Word M[$2+16] = $3
Op Ra Rb OffsetBranch
beq $3$2dest Branch when $3 = $2
9262016 (copyJP Shen) 18-600 Lecture 8 10
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Evolutions of MIPS and ARM ISAs
9262016 (copyJP Shen) 18-600 Lecture 8 11
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ISA Design Summary
How Important is ISA Design Less now than before
With enough hardware can make almost anything go fast Installed based of legacy code and SW development tools dominate
Y86-64 Instruction Set Architecture Similar state and instructions as x86-64 Simpler encodings Somewhere between CISC and RISC
9262016 (copyJP Shen) 18-600 Lecture 8 12
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Processor Statebull Program Registers
bull 15 registers (omit r15) Each 64 bits
bull Condition Codesbull Single-bit flags set by arithmetic or logical instructions
bull ZF Zero SFNegative OF Overflow
bull Program Counterbull Indicates address of next instruction
bull Program Statusbull Indicates either normal operation or some error condition
bull Memorybull Byte-addressable storage arraybull Words stored in little-endian byte order
ZFSFOF
RF Program registers
CC Condition
codes
PC
DMEM Memory
Stat Program status
r8
r9
r10
r11
r12
r13
r14
rax
rcx
rdx
rbx
rsp
rbp
rsi
rdi
9262016 (copyJP Shen) 18-600 Lecture 8 13
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Set
18-600 Lecture 89262016 (copyJP Shen) 14
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Formats
Formatbull 1ndash10 bytes of information read from memory
bull Can determine instruction length from first bytebull Fewer instruction types and simpler encoding than x86-64
bull Each instruction accesses and modifies some part(s) of the program state
9262016 (copyJP Shen) 18-600 Lecture 8 15
18-600 Lecture 89262016 (copyJP Shen) 16
0 1 2 3 4 5 6 7 8 9
V
D
D
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB
rmmovq rA D(rB) 4 0 rA rB
mrmovq D(rB) rA 5 0 rA rB
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
rrmovq 2 0
cmovle 2 1
cmovl 2 2
cmove 2 3
cmovne 2 4
cmovge 2 5
cmovg 2 6
Y86-64 Instructions (moves)
18-600 Lecture 89262016 (copyJP Shen) 17
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
addq 6 0
subq 6 1
andq 6 2
xorq 6 3
Y86-64 Instructions (ALUs)
18-600 Lecture 89262016 (copyJP Shen) 18
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9jmp 7 0
jle 7 1
jl 7 2
je 7 3
jne 7 4
jge 7 5
jg 7 6
Y86-64 Instructions (branches)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Encoding Register Specifiers
Each register has 4-bit ID
bull Same encoding as in x86-64 Register ID 15 (0xF) indicates ldquono registerrdquo
bull Will use this in our hardware design in multiple places
rax
rcx
rdx
rbx
0
1
2
3
rsp
rbp
rsi
rdi
4
5
6
7
r8
r9
r10
r11
8
9
A
B
r12
r13
r14
No Register
C
D
E
F
9262016 (copyJP Shen) 18-600 Lecture 8 19
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Format ExampleAddition Instruction
Add value in register rA to that in register rB Store result in register rB Note that Y86-64 only allows addition to be applied to register data
Set condition codes based on result eg addq raxrsi Encoding 60 06
Two-byte encoding First indicates instruction type Second gives source and destination registers
addq rA rB 6 0 rA rB
Encoded Representation
Generic Form
9262016 (copyJP Shen) 18-600 Lecture 8 20
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Arithmetic and Logical Operations
Refer to generically as ldquoOPqrdquo Encodings differ only by ldquofunction coderdquo
Low-order 4 bits in first instruction byte
Set condition codes as side effect
addq rA rB 6 0 rA rB
subq rA rB 6 1 rA rB
andq rA rB 6 2 rA rB
xorq rA rB 6 3 rA rB
Add
Subtract (rA from rB)
And
Exclusive-Or
Instruction Code Function Code
9262016 (copyJP Shen) 18-600 Lecture 8 21
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Operations
Like the x86-64 movq instruction Simpler format for memory addresses Give different names to keep them distinct
rrmovq rA rB 2 0
Register Register
Immediate Register
irmovq V rB F rB3 0 V
Register Memory
rmmovq rA D(rB) 4 0 rA rB D
Memory Register
mrmovq D(rB) rA 5 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 22
rA rB
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ComputerEngineering
Application Softwarebull Program developmentbull Program compilation
Hardware Technologybull Program executionbull Computer performance
Computer Instruction Set Architecture
9262016 (copyJP Shen) 18-600 Lecture 8 3
Instruction Set Architecture (ISA) SPECIFICATION
IMPLEMENTATION
SoftwareEngineering
Proc
esso
rDe
sign
Prog
ram
Deve
lopm
ent
CS APP
CS AAP
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Set Architecture Assembly Language View
bull Processor state (architecture state)bull Registers Memory hellip
bull Instructions (specify operations operands)bull addq pushq ret hellipbull How instructions are encoded as bytes
Layer of Abstractionbull Above how to program machine
bull Processor executes instructions in a sequencebull Below what needs to be implemented
bull Use variety of uarch techniques to make it run fastbull Eg execute multiple instructions simultaneously
ISA
Compiler OS
CPUDesign
CircuitDesign
ChipLayout
ApplicationProgram
9262016 (copyJP Shen) 18-600 Lecture 8 4
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CISC Instruction Sets Complex Instruction Set Computer IA32 is example
Stack-oriented instruction set Use stack to pass arguments save program counter Explicit push and pop instructions
Arithmetic instructions can access memory addq rax 12(rbxrcx8)
requires memory read and write Complex address calculation
Condition codes Set as side effect of arithmetic and logical instructions
Philosophy Add instructions to perform ldquotypicalrdquo programming tasks
9262016 (copyJP Shen) 18-600 Lecture 8 5
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
RISC Instruction Sets Reduced Instruction Set Computer Internal project at IBM later popularized by Hennessy (Stanford) and Patterson (Berkeley)
Fewer simpler instructions Might take more to get given task done Can execute them with small and fast hardware
Register-oriented instruction set Many more (typically 32) registers Use for arguments return pointer temporaries
Only load and store instructions can access memory Similar to Y86-64 mrmovq and rmmovq
No Condition codes Test instructions return 01 in register
9262016 (copyJP Shen) 18-600 Lecture 8 6
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Quantum Physics
Transistors amp Devices
Logic Gates amp Memory
Von Neumann Machine
x86 Machine Primitives
Visual C++
Firefox MS Excel
Windows
Computer Dynamic-Static Interface
9262016 (copyJP Shen) 18-600 Lecture 8 7
DynamicStatic Interface (DSI)=(ISA)
ldquosta
ticrdquo
ldquodyn
amic
rdquo
Architectural State
Microarchitecture State
MACHINE
PROGRAM
Exposed to SW
Hidden in HW
Buffering needed between arch and uarch statesbull Allow uarch state to deviate from arch statebull Able to undo speculative uarch state if needed
Architectural state requirementsbull Support sequential instruction execution semanticsbull Support precise servicing of exceptions amp interrupts
ARCHITECTURE
DSI = ISA = a contract between the program and the machine
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CISC vs RISCOriginal Debate
Strong opinions CISC proponents---easy for compiler fewer code bytes RISC proponents---better for optimizing compilers can make machine run fast with
simple chip design
Current Status For desktop processors choice of ISA not a technical issue
With enough hardware can make anything run fast Code compatibility more important
x86-64 adopted many RISC features More registers use them for argument passing
For embedded processors RISC makes sense Smaller cheaper less power Most mobile phones use ARM processors
9262016 (copyJP Shen) 18-600 Lecture 8 8
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
MIPS Registers$0$1$2$3$4$5$6$7$8$9$10$11$12$13$14$15
$0$at$v0$v1$a0$a1$a2$a3$t0$t1$t2$t3$t4$t5$t6$t7
Constant 0Reserved Temp
Return Values
Procedure arguments
Caller SaveTemporariesMay be overwritten bycalled procedures
$16$17$18$19$20$21$22$23$24$25$26$27$28$29$30$31
$s0$s1$s2$s3$s4$s5$s6$s7$t8$t9$k0$k1$gp$sp$s8$ra
Reserved forOperating Sys
Caller Save Temp
Global Pointer
Callee SaveTemporariesMay not beoverwritten bycalled procedures
Stack PointerCallee Save TempReturn Address
9262016 (copyJP Shen) 18-600 Lecture 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
MIPS Instruction Examples
Op Ra Rb Offset
Op Ra Rb Rd Fn00000R-R
Op Ra Rb ImmediateR-I
LoadStore
addu $3$2$1 Register add $3 = $2+$1
addu $3$2 3145 Immediate add $3 = $2+3145
sll $3$22 Shift left $3 = $2 ltlt 2
lw $316($2) Load Word $3 = M[$2+16]
sw $316($2) Store Word M[$2+16] = $3
Op Ra Rb OffsetBranch
beq $3$2dest Branch when $3 = $2
9262016 (copyJP Shen) 18-600 Lecture 8 10
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Evolutions of MIPS and ARM ISAs
9262016 (copyJP Shen) 18-600 Lecture 8 11
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ISA Design Summary
How Important is ISA Design Less now than before
With enough hardware can make almost anything go fast Installed based of legacy code and SW development tools dominate
Y86-64 Instruction Set Architecture Similar state and instructions as x86-64 Simpler encodings Somewhere between CISC and RISC
9262016 (copyJP Shen) 18-600 Lecture 8 12
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Processor Statebull Program Registers
bull 15 registers (omit r15) Each 64 bits
bull Condition Codesbull Single-bit flags set by arithmetic or logical instructions
bull ZF Zero SFNegative OF Overflow
bull Program Counterbull Indicates address of next instruction
bull Program Statusbull Indicates either normal operation or some error condition
bull Memorybull Byte-addressable storage arraybull Words stored in little-endian byte order
ZFSFOF
RF Program registers
CC Condition
codes
PC
DMEM Memory
Stat Program status
r8
r9
r10
r11
r12
r13
r14
rax
rcx
rdx
rbx
rsp
rbp
rsi
rdi
9262016 (copyJP Shen) 18-600 Lecture 8 13
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Set
18-600 Lecture 89262016 (copyJP Shen) 14
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Formats
Formatbull 1ndash10 bytes of information read from memory
bull Can determine instruction length from first bytebull Fewer instruction types and simpler encoding than x86-64
bull Each instruction accesses and modifies some part(s) of the program state
9262016 (copyJP Shen) 18-600 Lecture 8 15
18-600 Lecture 89262016 (copyJP Shen) 16
0 1 2 3 4 5 6 7 8 9
V
D
D
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB
rmmovq rA D(rB) 4 0 rA rB
mrmovq D(rB) rA 5 0 rA rB
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
rrmovq 2 0
cmovle 2 1
cmovl 2 2
cmove 2 3
cmovne 2 4
cmovge 2 5
cmovg 2 6
Y86-64 Instructions (moves)
18-600 Lecture 89262016 (copyJP Shen) 17
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
addq 6 0
subq 6 1
andq 6 2
xorq 6 3
Y86-64 Instructions (ALUs)
18-600 Lecture 89262016 (copyJP Shen) 18
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9jmp 7 0
jle 7 1
jl 7 2
je 7 3
jne 7 4
jge 7 5
jg 7 6
Y86-64 Instructions (branches)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Encoding Register Specifiers
Each register has 4-bit ID
bull Same encoding as in x86-64 Register ID 15 (0xF) indicates ldquono registerrdquo
bull Will use this in our hardware design in multiple places
rax
rcx
rdx
rbx
0
1
2
3
rsp
rbp
rsi
rdi
4
5
6
7
r8
r9
r10
r11
8
9
A
B
r12
r13
r14
No Register
C
D
E
F
9262016 (copyJP Shen) 18-600 Lecture 8 19
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Format ExampleAddition Instruction
Add value in register rA to that in register rB Store result in register rB Note that Y86-64 only allows addition to be applied to register data
Set condition codes based on result eg addq raxrsi Encoding 60 06
Two-byte encoding First indicates instruction type Second gives source and destination registers
addq rA rB 6 0 rA rB
Encoded Representation
Generic Form
9262016 (copyJP Shen) 18-600 Lecture 8 20
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Arithmetic and Logical Operations
Refer to generically as ldquoOPqrdquo Encodings differ only by ldquofunction coderdquo
Low-order 4 bits in first instruction byte
Set condition codes as side effect
addq rA rB 6 0 rA rB
subq rA rB 6 1 rA rB
andq rA rB 6 2 rA rB
xorq rA rB 6 3 rA rB
Add
Subtract (rA from rB)
And
Exclusive-Or
Instruction Code Function Code
9262016 (copyJP Shen) 18-600 Lecture 8 21
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Operations
Like the x86-64 movq instruction Simpler format for memory addresses Give different names to keep them distinct
rrmovq rA rB 2 0
Register Register
Immediate Register
irmovq V rB F rB3 0 V
Register Memory
rmmovq rA D(rB) 4 0 rA rB D
Memory Register
mrmovq D(rB) rA 5 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 22
rA rB
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Set Architecture Assembly Language View
bull Processor state (architecture state)bull Registers Memory hellip
bull Instructions (specify operations operands)bull addq pushq ret hellipbull How instructions are encoded as bytes
Layer of Abstractionbull Above how to program machine
bull Processor executes instructions in a sequencebull Below what needs to be implemented
bull Use variety of uarch techniques to make it run fastbull Eg execute multiple instructions simultaneously
ISA
Compiler OS
CPUDesign
CircuitDesign
ChipLayout
ApplicationProgram
9262016 (copyJP Shen) 18-600 Lecture 8 4
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CISC Instruction Sets Complex Instruction Set Computer IA32 is example
Stack-oriented instruction set Use stack to pass arguments save program counter Explicit push and pop instructions
Arithmetic instructions can access memory addq rax 12(rbxrcx8)
requires memory read and write Complex address calculation
Condition codes Set as side effect of arithmetic and logical instructions
Philosophy Add instructions to perform ldquotypicalrdquo programming tasks
9262016 (copyJP Shen) 18-600 Lecture 8 5
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
RISC Instruction Sets Reduced Instruction Set Computer Internal project at IBM later popularized by Hennessy (Stanford) and Patterson (Berkeley)
Fewer simpler instructions Might take more to get given task done Can execute them with small and fast hardware
Register-oriented instruction set Many more (typically 32) registers Use for arguments return pointer temporaries
Only load and store instructions can access memory Similar to Y86-64 mrmovq and rmmovq
No Condition codes Test instructions return 01 in register
9262016 (copyJP Shen) 18-600 Lecture 8 6
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Quantum Physics
Transistors amp Devices
Logic Gates amp Memory
Von Neumann Machine
x86 Machine Primitives
Visual C++
Firefox MS Excel
Windows
Computer Dynamic-Static Interface
9262016 (copyJP Shen) 18-600 Lecture 8 7
DynamicStatic Interface (DSI)=(ISA)
ldquosta
ticrdquo
ldquodyn
amic
rdquo
Architectural State
Microarchitecture State
MACHINE
PROGRAM
Exposed to SW
Hidden in HW
Buffering needed between arch and uarch statesbull Allow uarch state to deviate from arch statebull Able to undo speculative uarch state if needed
Architectural state requirementsbull Support sequential instruction execution semanticsbull Support precise servicing of exceptions amp interrupts
ARCHITECTURE
DSI = ISA = a contract between the program and the machine
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CISC vs RISCOriginal Debate
Strong opinions CISC proponents---easy for compiler fewer code bytes RISC proponents---better for optimizing compilers can make machine run fast with
simple chip design
Current Status For desktop processors choice of ISA not a technical issue
With enough hardware can make anything run fast Code compatibility more important
x86-64 adopted many RISC features More registers use them for argument passing
For embedded processors RISC makes sense Smaller cheaper less power Most mobile phones use ARM processors
9262016 (copyJP Shen) 18-600 Lecture 8 8
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
MIPS Registers$0$1$2$3$4$5$6$7$8$9$10$11$12$13$14$15
$0$at$v0$v1$a0$a1$a2$a3$t0$t1$t2$t3$t4$t5$t6$t7
Constant 0Reserved Temp
Return Values
Procedure arguments
Caller SaveTemporariesMay be overwritten bycalled procedures
$16$17$18$19$20$21$22$23$24$25$26$27$28$29$30$31
$s0$s1$s2$s3$s4$s5$s6$s7$t8$t9$k0$k1$gp$sp$s8$ra
Reserved forOperating Sys
Caller Save Temp
Global Pointer
Callee SaveTemporariesMay not beoverwritten bycalled procedures
Stack PointerCallee Save TempReturn Address
9262016 (copyJP Shen) 18-600 Lecture 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
MIPS Instruction Examples
Op Ra Rb Offset
Op Ra Rb Rd Fn00000R-R
Op Ra Rb ImmediateR-I
LoadStore
addu $3$2$1 Register add $3 = $2+$1
addu $3$2 3145 Immediate add $3 = $2+3145
sll $3$22 Shift left $3 = $2 ltlt 2
lw $316($2) Load Word $3 = M[$2+16]
sw $316($2) Store Word M[$2+16] = $3
Op Ra Rb OffsetBranch
beq $3$2dest Branch when $3 = $2
9262016 (copyJP Shen) 18-600 Lecture 8 10
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Evolutions of MIPS and ARM ISAs
9262016 (copyJP Shen) 18-600 Lecture 8 11
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ISA Design Summary
How Important is ISA Design Less now than before
With enough hardware can make almost anything go fast Installed based of legacy code and SW development tools dominate
Y86-64 Instruction Set Architecture Similar state and instructions as x86-64 Simpler encodings Somewhere between CISC and RISC
9262016 (copyJP Shen) 18-600 Lecture 8 12
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Processor Statebull Program Registers
bull 15 registers (omit r15) Each 64 bits
bull Condition Codesbull Single-bit flags set by arithmetic or logical instructions
bull ZF Zero SFNegative OF Overflow
bull Program Counterbull Indicates address of next instruction
bull Program Statusbull Indicates either normal operation or some error condition
bull Memorybull Byte-addressable storage arraybull Words stored in little-endian byte order
ZFSFOF
RF Program registers
CC Condition
codes
PC
DMEM Memory
Stat Program status
r8
r9
r10
r11
r12
r13
r14
rax
rcx
rdx
rbx
rsp
rbp
rsi
rdi
9262016 (copyJP Shen) 18-600 Lecture 8 13
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Set
18-600 Lecture 89262016 (copyJP Shen) 14
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Formats
Formatbull 1ndash10 bytes of information read from memory
bull Can determine instruction length from first bytebull Fewer instruction types and simpler encoding than x86-64
bull Each instruction accesses and modifies some part(s) of the program state
9262016 (copyJP Shen) 18-600 Lecture 8 15
18-600 Lecture 89262016 (copyJP Shen) 16
0 1 2 3 4 5 6 7 8 9
V
D
D
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB
rmmovq rA D(rB) 4 0 rA rB
mrmovq D(rB) rA 5 0 rA rB
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
rrmovq 2 0
cmovle 2 1
cmovl 2 2
cmove 2 3
cmovne 2 4
cmovge 2 5
cmovg 2 6
Y86-64 Instructions (moves)
18-600 Lecture 89262016 (copyJP Shen) 17
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
addq 6 0
subq 6 1
andq 6 2
xorq 6 3
Y86-64 Instructions (ALUs)
18-600 Lecture 89262016 (copyJP Shen) 18
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9jmp 7 0
jle 7 1
jl 7 2
je 7 3
jne 7 4
jge 7 5
jg 7 6
Y86-64 Instructions (branches)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Encoding Register Specifiers
Each register has 4-bit ID
bull Same encoding as in x86-64 Register ID 15 (0xF) indicates ldquono registerrdquo
bull Will use this in our hardware design in multiple places
rax
rcx
rdx
rbx
0
1
2
3
rsp
rbp
rsi
rdi
4
5
6
7
r8
r9
r10
r11
8
9
A
B
r12
r13
r14
No Register
C
D
E
F
9262016 (copyJP Shen) 18-600 Lecture 8 19
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Format ExampleAddition Instruction
Add value in register rA to that in register rB Store result in register rB Note that Y86-64 only allows addition to be applied to register data
Set condition codes based on result eg addq raxrsi Encoding 60 06
Two-byte encoding First indicates instruction type Second gives source and destination registers
addq rA rB 6 0 rA rB
Encoded Representation
Generic Form
9262016 (copyJP Shen) 18-600 Lecture 8 20
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Arithmetic and Logical Operations
Refer to generically as ldquoOPqrdquo Encodings differ only by ldquofunction coderdquo
Low-order 4 bits in first instruction byte
Set condition codes as side effect
addq rA rB 6 0 rA rB
subq rA rB 6 1 rA rB
andq rA rB 6 2 rA rB
xorq rA rB 6 3 rA rB
Add
Subtract (rA from rB)
And
Exclusive-Or
Instruction Code Function Code
9262016 (copyJP Shen) 18-600 Lecture 8 21
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Operations
Like the x86-64 movq instruction Simpler format for memory addresses Give different names to keep them distinct
rrmovq rA rB 2 0
Register Register
Immediate Register
irmovq V rB F rB3 0 V
Register Memory
rmmovq rA D(rB) 4 0 rA rB D
Memory Register
mrmovq D(rB) rA 5 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 22
rA rB
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CISC Instruction Sets Complex Instruction Set Computer IA32 is example
Stack-oriented instruction set Use stack to pass arguments save program counter Explicit push and pop instructions
Arithmetic instructions can access memory addq rax 12(rbxrcx8)
requires memory read and write Complex address calculation
Condition codes Set as side effect of arithmetic and logical instructions
Philosophy Add instructions to perform ldquotypicalrdquo programming tasks
9262016 (copyJP Shen) 18-600 Lecture 8 5
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
RISC Instruction Sets Reduced Instruction Set Computer Internal project at IBM later popularized by Hennessy (Stanford) and Patterson (Berkeley)
Fewer simpler instructions Might take more to get given task done Can execute them with small and fast hardware
Register-oriented instruction set Many more (typically 32) registers Use for arguments return pointer temporaries
Only load and store instructions can access memory Similar to Y86-64 mrmovq and rmmovq
No Condition codes Test instructions return 01 in register
9262016 (copyJP Shen) 18-600 Lecture 8 6
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Quantum Physics
Transistors amp Devices
Logic Gates amp Memory
Von Neumann Machine
x86 Machine Primitives
Visual C++
Firefox MS Excel
Windows
Computer Dynamic-Static Interface
9262016 (copyJP Shen) 18-600 Lecture 8 7
DynamicStatic Interface (DSI)=(ISA)
ldquosta
ticrdquo
ldquodyn
amic
rdquo
Architectural State
Microarchitecture State
MACHINE
PROGRAM
Exposed to SW
Hidden in HW
Buffering needed between arch and uarch statesbull Allow uarch state to deviate from arch statebull Able to undo speculative uarch state if needed
Architectural state requirementsbull Support sequential instruction execution semanticsbull Support precise servicing of exceptions amp interrupts
ARCHITECTURE
DSI = ISA = a contract between the program and the machine
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CISC vs RISCOriginal Debate
Strong opinions CISC proponents---easy for compiler fewer code bytes RISC proponents---better for optimizing compilers can make machine run fast with
simple chip design
Current Status For desktop processors choice of ISA not a technical issue
With enough hardware can make anything run fast Code compatibility more important
x86-64 adopted many RISC features More registers use them for argument passing
For embedded processors RISC makes sense Smaller cheaper less power Most mobile phones use ARM processors
9262016 (copyJP Shen) 18-600 Lecture 8 8
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
MIPS Registers$0$1$2$3$4$5$6$7$8$9$10$11$12$13$14$15
$0$at$v0$v1$a0$a1$a2$a3$t0$t1$t2$t3$t4$t5$t6$t7
Constant 0Reserved Temp
Return Values
Procedure arguments
Caller SaveTemporariesMay be overwritten bycalled procedures
$16$17$18$19$20$21$22$23$24$25$26$27$28$29$30$31
$s0$s1$s2$s3$s4$s5$s6$s7$t8$t9$k0$k1$gp$sp$s8$ra
Reserved forOperating Sys
Caller Save Temp
Global Pointer
Callee SaveTemporariesMay not beoverwritten bycalled procedures
Stack PointerCallee Save TempReturn Address
9262016 (copyJP Shen) 18-600 Lecture 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
MIPS Instruction Examples
Op Ra Rb Offset
Op Ra Rb Rd Fn00000R-R
Op Ra Rb ImmediateR-I
LoadStore
addu $3$2$1 Register add $3 = $2+$1
addu $3$2 3145 Immediate add $3 = $2+3145
sll $3$22 Shift left $3 = $2 ltlt 2
lw $316($2) Load Word $3 = M[$2+16]
sw $316($2) Store Word M[$2+16] = $3
Op Ra Rb OffsetBranch
beq $3$2dest Branch when $3 = $2
9262016 (copyJP Shen) 18-600 Lecture 8 10
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Evolutions of MIPS and ARM ISAs
9262016 (copyJP Shen) 18-600 Lecture 8 11
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ISA Design Summary
How Important is ISA Design Less now than before
With enough hardware can make almost anything go fast Installed based of legacy code and SW development tools dominate
Y86-64 Instruction Set Architecture Similar state and instructions as x86-64 Simpler encodings Somewhere between CISC and RISC
9262016 (copyJP Shen) 18-600 Lecture 8 12
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Processor Statebull Program Registers
bull 15 registers (omit r15) Each 64 bits
bull Condition Codesbull Single-bit flags set by arithmetic or logical instructions
bull ZF Zero SFNegative OF Overflow
bull Program Counterbull Indicates address of next instruction
bull Program Statusbull Indicates either normal operation or some error condition
bull Memorybull Byte-addressable storage arraybull Words stored in little-endian byte order
ZFSFOF
RF Program registers
CC Condition
codes
PC
DMEM Memory
Stat Program status
r8
r9
r10
r11
r12
r13
r14
rax
rcx
rdx
rbx
rsp
rbp
rsi
rdi
9262016 (copyJP Shen) 18-600 Lecture 8 13
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Set
18-600 Lecture 89262016 (copyJP Shen) 14
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Formats
Formatbull 1ndash10 bytes of information read from memory
bull Can determine instruction length from first bytebull Fewer instruction types and simpler encoding than x86-64
bull Each instruction accesses and modifies some part(s) of the program state
9262016 (copyJP Shen) 18-600 Lecture 8 15
18-600 Lecture 89262016 (copyJP Shen) 16
0 1 2 3 4 5 6 7 8 9
V
D
D
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB
rmmovq rA D(rB) 4 0 rA rB
mrmovq D(rB) rA 5 0 rA rB
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
rrmovq 2 0
cmovle 2 1
cmovl 2 2
cmove 2 3
cmovne 2 4
cmovge 2 5
cmovg 2 6
Y86-64 Instructions (moves)
18-600 Lecture 89262016 (copyJP Shen) 17
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
addq 6 0
subq 6 1
andq 6 2
xorq 6 3
Y86-64 Instructions (ALUs)
18-600 Lecture 89262016 (copyJP Shen) 18
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9jmp 7 0
jle 7 1
jl 7 2
je 7 3
jne 7 4
jge 7 5
jg 7 6
Y86-64 Instructions (branches)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Encoding Register Specifiers
Each register has 4-bit ID
bull Same encoding as in x86-64 Register ID 15 (0xF) indicates ldquono registerrdquo
bull Will use this in our hardware design in multiple places
rax
rcx
rdx
rbx
0
1
2
3
rsp
rbp
rsi
rdi
4
5
6
7
r8
r9
r10
r11
8
9
A
B
r12
r13
r14
No Register
C
D
E
F
9262016 (copyJP Shen) 18-600 Lecture 8 19
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Format ExampleAddition Instruction
Add value in register rA to that in register rB Store result in register rB Note that Y86-64 only allows addition to be applied to register data
Set condition codes based on result eg addq raxrsi Encoding 60 06
Two-byte encoding First indicates instruction type Second gives source and destination registers
addq rA rB 6 0 rA rB
Encoded Representation
Generic Form
9262016 (copyJP Shen) 18-600 Lecture 8 20
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Arithmetic and Logical Operations
Refer to generically as ldquoOPqrdquo Encodings differ only by ldquofunction coderdquo
Low-order 4 bits in first instruction byte
Set condition codes as side effect
addq rA rB 6 0 rA rB
subq rA rB 6 1 rA rB
andq rA rB 6 2 rA rB
xorq rA rB 6 3 rA rB
Add
Subtract (rA from rB)
And
Exclusive-Or
Instruction Code Function Code
9262016 (copyJP Shen) 18-600 Lecture 8 21
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Operations
Like the x86-64 movq instruction Simpler format for memory addresses Give different names to keep them distinct
rrmovq rA rB 2 0
Register Register
Immediate Register
irmovq V rB F rB3 0 V
Register Memory
rmmovq rA D(rB) 4 0 rA rB D
Memory Register
mrmovq D(rB) rA 5 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 22
rA rB
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
RISC Instruction Sets Reduced Instruction Set Computer Internal project at IBM later popularized by Hennessy (Stanford) and Patterson (Berkeley)
Fewer simpler instructions Might take more to get given task done Can execute them with small and fast hardware
Register-oriented instruction set Many more (typically 32) registers Use for arguments return pointer temporaries
Only load and store instructions can access memory Similar to Y86-64 mrmovq and rmmovq
No Condition codes Test instructions return 01 in register
9262016 (copyJP Shen) 18-600 Lecture 8 6
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Quantum Physics
Transistors amp Devices
Logic Gates amp Memory
Von Neumann Machine
x86 Machine Primitives
Visual C++
Firefox MS Excel
Windows
Computer Dynamic-Static Interface
9262016 (copyJP Shen) 18-600 Lecture 8 7
DynamicStatic Interface (DSI)=(ISA)
ldquosta
ticrdquo
ldquodyn
amic
rdquo
Architectural State
Microarchitecture State
MACHINE
PROGRAM
Exposed to SW
Hidden in HW
Buffering needed between arch and uarch statesbull Allow uarch state to deviate from arch statebull Able to undo speculative uarch state if needed
Architectural state requirementsbull Support sequential instruction execution semanticsbull Support precise servicing of exceptions amp interrupts
ARCHITECTURE
DSI = ISA = a contract between the program and the machine
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CISC vs RISCOriginal Debate
Strong opinions CISC proponents---easy for compiler fewer code bytes RISC proponents---better for optimizing compilers can make machine run fast with
simple chip design
Current Status For desktop processors choice of ISA not a technical issue
With enough hardware can make anything run fast Code compatibility more important
x86-64 adopted many RISC features More registers use them for argument passing
For embedded processors RISC makes sense Smaller cheaper less power Most mobile phones use ARM processors
9262016 (copyJP Shen) 18-600 Lecture 8 8
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
MIPS Registers$0$1$2$3$4$5$6$7$8$9$10$11$12$13$14$15
$0$at$v0$v1$a0$a1$a2$a3$t0$t1$t2$t3$t4$t5$t6$t7
Constant 0Reserved Temp
Return Values
Procedure arguments
Caller SaveTemporariesMay be overwritten bycalled procedures
$16$17$18$19$20$21$22$23$24$25$26$27$28$29$30$31
$s0$s1$s2$s3$s4$s5$s6$s7$t8$t9$k0$k1$gp$sp$s8$ra
Reserved forOperating Sys
Caller Save Temp
Global Pointer
Callee SaveTemporariesMay not beoverwritten bycalled procedures
Stack PointerCallee Save TempReturn Address
9262016 (copyJP Shen) 18-600 Lecture 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
MIPS Instruction Examples
Op Ra Rb Offset
Op Ra Rb Rd Fn00000R-R
Op Ra Rb ImmediateR-I
LoadStore
addu $3$2$1 Register add $3 = $2+$1
addu $3$2 3145 Immediate add $3 = $2+3145
sll $3$22 Shift left $3 = $2 ltlt 2
lw $316($2) Load Word $3 = M[$2+16]
sw $316($2) Store Word M[$2+16] = $3
Op Ra Rb OffsetBranch
beq $3$2dest Branch when $3 = $2
9262016 (copyJP Shen) 18-600 Lecture 8 10
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Evolutions of MIPS and ARM ISAs
9262016 (copyJP Shen) 18-600 Lecture 8 11
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ISA Design Summary
How Important is ISA Design Less now than before
With enough hardware can make almost anything go fast Installed based of legacy code and SW development tools dominate
Y86-64 Instruction Set Architecture Similar state and instructions as x86-64 Simpler encodings Somewhere between CISC and RISC
9262016 (copyJP Shen) 18-600 Lecture 8 12
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Processor Statebull Program Registers
bull 15 registers (omit r15) Each 64 bits
bull Condition Codesbull Single-bit flags set by arithmetic or logical instructions
bull ZF Zero SFNegative OF Overflow
bull Program Counterbull Indicates address of next instruction
bull Program Statusbull Indicates either normal operation or some error condition
bull Memorybull Byte-addressable storage arraybull Words stored in little-endian byte order
ZFSFOF
RF Program registers
CC Condition
codes
PC
DMEM Memory
Stat Program status
r8
r9
r10
r11
r12
r13
r14
rax
rcx
rdx
rbx
rsp
rbp
rsi
rdi
9262016 (copyJP Shen) 18-600 Lecture 8 13
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Set
18-600 Lecture 89262016 (copyJP Shen) 14
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Formats
Formatbull 1ndash10 bytes of information read from memory
bull Can determine instruction length from first bytebull Fewer instruction types and simpler encoding than x86-64
bull Each instruction accesses and modifies some part(s) of the program state
9262016 (copyJP Shen) 18-600 Lecture 8 15
18-600 Lecture 89262016 (copyJP Shen) 16
0 1 2 3 4 5 6 7 8 9
V
D
D
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB
rmmovq rA D(rB) 4 0 rA rB
mrmovq D(rB) rA 5 0 rA rB
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
rrmovq 2 0
cmovle 2 1
cmovl 2 2
cmove 2 3
cmovne 2 4
cmovge 2 5
cmovg 2 6
Y86-64 Instructions (moves)
18-600 Lecture 89262016 (copyJP Shen) 17
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
addq 6 0
subq 6 1
andq 6 2
xorq 6 3
Y86-64 Instructions (ALUs)
18-600 Lecture 89262016 (copyJP Shen) 18
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9jmp 7 0
jle 7 1
jl 7 2
je 7 3
jne 7 4
jge 7 5
jg 7 6
Y86-64 Instructions (branches)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Encoding Register Specifiers
Each register has 4-bit ID
bull Same encoding as in x86-64 Register ID 15 (0xF) indicates ldquono registerrdquo
bull Will use this in our hardware design in multiple places
rax
rcx
rdx
rbx
0
1
2
3
rsp
rbp
rsi
rdi
4
5
6
7
r8
r9
r10
r11
8
9
A
B
r12
r13
r14
No Register
C
D
E
F
9262016 (copyJP Shen) 18-600 Lecture 8 19
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Format ExampleAddition Instruction
Add value in register rA to that in register rB Store result in register rB Note that Y86-64 only allows addition to be applied to register data
Set condition codes based on result eg addq raxrsi Encoding 60 06
Two-byte encoding First indicates instruction type Second gives source and destination registers
addq rA rB 6 0 rA rB
Encoded Representation
Generic Form
9262016 (copyJP Shen) 18-600 Lecture 8 20
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Arithmetic and Logical Operations
Refer to generically as ldquoOPqrdquo Encodings differ only by ldquofunction coderdquo
Low-order 4 bits in first instruction byte
Set condition codes as side effect
addq rA rB 6 0 rA rB
subq rA rB 6 1 rA rB
andq rA rB 6 2 rA rB
xorq rA rB 6 3 rA rB
Add
Subtract (rA from rB)
And
Exclusive-Or
Instruction Code Function Code
9262016 (copyJP Shen) 18-600 Lecture 8 21
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Operations
Like the x86-64 movq instruction Simpler format for memory addresses Give different names to keep them distinct
rrmovq rA rB 2 0
Register Register
Immediate Register
irmovq V rB F rB3 0 V
Register Memory
rmmovq rA D(rB) 4 0 rA rB D
Memory Register
mrmovq D(rB) rA 5 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 22
rA rB
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Quantum Physics
Transistors amp Devices
Logic Gates amp Memory
Von Neumann Machine
x86 Machine Primitives
Visual C++
Firefox MS Excel
Windows
Computer Dynamic-Static Interface
9262016 (copyJP Shen) 18-600 Lecture 8 7
DynamicStatic Interface (DSI)=(ISA)
ldquosta
ticrdquo
ldquodyn
amic
rdquo
Architectural State
Microarchitecture State
MACHINE
PROGRAM
Exposed to SW
Hidden in HW
Buffering needed between arch and uarch statesbull Allow uarch state to deviate from arch statebull Able to undo speculative uarch state if needed
Architectural state requirementsbull Support sequential instruction execution semanticsbull Support precise servicing of exceptions amp interrupts
ARCHITECTURE
DSI = ISA = a contract between the program and the machine
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CISC vs RISCOriginal Debate
Strong opinions CISC proponents---easy for compiler fewer code bytes RISC proponents---better for optimizing compilers can make machine run fast with
simple chip design
Current Status For desktop processors choice of ISA not a technical issue
With enough hardware can make anything run fast Code compatibility more important
x86-64 adopted many RISC features More registers use them for argument passing
For embedded processors RISC makes sense Smaller cheaper less power Most mobile phones use ARM processors
9262016 (copyJP Shen) 18-600 Lecture 8 8
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
MIPS Registers$0$1$2$3$4$5$6$7$8$9$10$11$12$13$14$15
$0$at$v0$v1$a0$a1$a2$a3$t0$t1$t2$t3$t4$t5$t6$t7
Constant 0Reserved Temp
Return Values
Procedure arguments
Caller SaveTemporariesMay be overwritten bycalled procedures
$16$17$18$19$20$21$22$23$24$25$26$27$28$29$30$31
$s0$s1$s2$s3$s4$s5$s6$s7$t8$t9$k0$k1$gp$sp$s8$ra
Reserved forOperating Sys
Caller Save Temp
Global Pointer
Callee SaveTemporariesMay not beoverwritten bycalled procedures
Stack PointerCallee Save TempReturn Address
9262016 (copyJP Shen) 18-600 Lecture 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
MIPS Instruction Examples
Op Ra Rb Offset
Op Ra Rb Rd Fn00000R-R
Op Ra Rb ImmediateR-I
LoadStore
addu $3$2$1 Register add $3 = $2+$1
addu $3$2 3145 Immediate add $3 = $2+3145
sll $3$22 Shift left $3 = $2 ltlt 2
lw $316($2) Load Word $3 = M[$2+16]
sw $316($2) Store Word M[$2+16] = $3
Op Ra Rb OffsetBranch
beq $3$2dest Branch when $3 = $2
9262016 (copyJP Shen) 18-600 Lecture 8 10
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Evolutions of MIPS and ARM ISAs
9262016 (copyJP Shen) 18-600 Lecture 8 11
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ISA Design Summary
How Important is ISA Design Less now than before
With enough hardware can make almost anything go fast Installed based of legacy code and SW development tools dominate
Y86-64 Instruction Set Architecture Similar state and instructions as x86-64 Simpler encodings Somewhere between CISC and RISC
9262016 (copyJP Shen) 18-600 Lecture 8 12
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Processor Statebull Program Registers
bull 15 registers (omit r15) Each 64 bits
bull Condition Codesbull Single-bit flags set by arithmetic or logical instructions
bull ZF Zero SFNegative OF Overflow
bull Program Counterbull Indicates address of next instruction
bull Program Statusbull Indicates either normal operation or some error condition
bull Memorybull Byte-addressable storage arraybull Words stored in little-endian byte order
ZFSFOF
RF Program registers
CC Condition
codes
PC
DMEM Memory
Stat Program status
r8
r9
r10
r11
r12
r13
r14
rax
rcx
rdx
rbx
rsp
rbp
rsi
rdi
9262016 (copyJP Shen) 18-600 Lecture 8 13
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Set
18-600 Lecture 89262016 (copyJP Shen) 14
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Formats
Formatbull 1ndash10 bytes of information read from memory
bull Can determine instruction length from first bytebull Fewer instruction types and simpler encoding than x86-64
bull Each instruction accesses and modifies some part(s) of the program state
9262016 (copyJP Shen) 18-600 Lecture 8 15
18-600 Lecture 89262016 (copyJP Shen) 16
0 1 2 3 4 5 6 7 8 9
V
D
D
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB
rmmovq rA D(rB) 4 0 rA rB
mrmovq D(rB) rA 5 0 rA rB
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
rrmovq 2 0
cmovle 2 1
cmovl 2 2
cmove 2 3
cmovne 2 4
cmovge 2 5
cmovg 2 6
Y86-64 Instructions (moves)
18-600 Lecture 89262016 (copyJP Shen) 17
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
addq 6 0
subq 6 1
andq 6 2
xorq 6 3
Y86-64 Instructions (ALUs)
18-600 Lecture 89262016 (copyJP Shen) 18
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9jmp 7 0
jle 7 1
jl 7 2
je 7 3
jne 7 4
jge 7 5
jg 7 6
Y86-64 Instructions (branches)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Encoding Register Specifiers
Each register has 4-bit ID
bull Same encoding as in x86-64 Register ID 15 (0xF) indicates ldquono registerrdquo
bull Will use this in our hardware design in multiple places
rax
rcx
rdx
rbx
0
1
2
3
rsp
rbp
rsi
rdi
4
5
6
7
r8
r9
r10
r11
8
9
A
B
r12
r13
r14
No Register
C
D
E
F
9262016 (copyJP Shen) 18-600 Lecture 8 19
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Format ExampleAddition Instruction
Add value in register rA to that in register rB Store result in register rB Note that Y86-64 only allows addition to be applied to register data
Set condition codes based on result eg addq raxrsi Encoding 60 06
Two-byte encoding First indicates instruction type Second gives source and destination registers
addq rA rB 6 0 rA rB
Encoded Representation
Generic Form
9262016 (copyJP Shen) 18-600 Lecture 8 20
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Arithmetic and Logical Operations
Refer to generically as ldquoOPqrdquo Encodings differ only by ldquofunction coderdquo
Low-order 4 bits in first instruction byte
Set condition codes as side effect
addq rA rB 6 0 rA rB
subq rA rB 6 1 rA rB
andq rA rB 6 2 rA rB
xorq rA rB 6 3 rA rB
Add
Subtract (rA from rB)
And
Exclusive-Or
Instruction Code Function Code
9262016 (copyJP Shen) 18-600 Lecture 8 21
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Operations
Like the x86-64 movq instruction Simpler format for memory addresses Give different names to keep them distinct
rrmovq rA rB 2 0
Register Register
Immediate Register
irmovq V rB F rB3 0 V
Register Memory
rmmovq rA D(rB) 4 0 rA rB D
Memory Register
mrmovq D(rB) rA 5 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 22
rA rB
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CISC vs RISCOriginal Debate
Strong opinions CISC proponents---easy for compiler fewer code bytes RISC proponents---better for optimizing compilers can make machine run fast with
simple chip design
Current Status For desktop processors choice of ISA not a technical issue
With enough hardware can make anything run fast Code compatibility more important
x86-64 adopted many RISC features More registers use them for argument passing
For embedded processors RISC makes sense Smaller cheaper less power Most mobile phones use ARM processors
9262016 (copyJP Shen) 18-600 Lecture 8 8
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
MIPS Registers$0$1$2$3$4$5$6$7$8$9$10$11$12$13$14$15
$0$at$v0$v1$a0$a1$a2$a3$t0$t1$t2$t3$t4$t5$t6$t7
Constant 0Reserved Temp
Return Values
Procedure arguments
Caller SaveTemporariesMay be overwritten bycalled procedures
$16$17$18$19$20$21$22$23$24$25$26$27$28$29$30$31
$s0$s1$s2$s3$s4$s5$s6$s7$t8$t9$k0$k1$gp$sp$s8$ra
Reserved forOperating Sys
Caller Save Temp
Global Pointer
Callee SaveTemporariesMay not beoverwritten bycalled procedures
Stack PointerCallee Save TempReturn Address
9262016 (copyJP Shen) 18-600 Lecture 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
MIPS Instruction Examples
Op Ra Rb Offset
Op Ra Rb Rd Fn00000R-R
Op Ra Rb ImmediateR-I
LoadStore
addu $3$2$1 Register add $3 = $2+$1
addu $3$2 3145 Immediate add $3 = $2+3145
sll $3$22 Shift left $3 = $2 ltlt 2
lw $316($2) Load Word $3 = M[$2+16]
sw $316($2) Store Word M[$2+16] = $3
Op Ra Rb OffsetBranch
beq $3$2dest Branch when $3 = $2
9262016 (copyJP Shen) 18-600 Lecture 8 10
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Evolutions of MIPS and ARM ISAs
9262016 (copyJP Shen) 18-600 Lecture 8 11
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ISA Design Summary
How Important is ISA Design Less now than before
With enough hardware can make almost anything go fast Installed based of legacy code and SW development tools dominate
Y86-64 Instruction Set Architecture Similar state and instructions as x86-64 Simpler encodings Somewhere between CISC and RISC
9262016 (copyJP Shen) 18-600 Lecture 8 12
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Processor Statebull Program Registers
bull 15 registers (omit r15) Each 64 bits
bull Condition Codesbull Single-bit flags set by arithmetic or logical instructions
bull ZF Zero SFNegative OF Overflow
bull Program Counterbull Indicates address of next instruction
bull Program Statusbull Indicates either normal operation or some error condition
bull Memorybull Byte-addressable storage arraybull Words stored in little-endian byte order
ZFSFOF
RF Program registers
CC Condition
codes
PC
DMEM Memory
Stat Program status
r8
r9
r10
r11
r12
r13
r14
rax
rcx
rdx
rbx
rsp
rbp
rsi
rdi
9262016 (copyJP Shen) 18-600 Lecture 8 13
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Set
18-600 Lecture 89262016 (copyJP Shen) 14
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Formats
Formatbull 1ndash10 bytes of information read from memory
bull Can determine instruction length from first bytebull Fewer instruction types and simpler encoding than x86-64
bull Each instruction accesses and modifies some part(s) of the program state
9262016 (copyJP Shen) 18-600 Lecture 8 15
18-600 Lecture 89262016 (copyJP Shen) 16
0 1 2 3 4 5 6 7 8 9
V
D
D
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB
rmmovq rA D(rB) 4 0 rA rB
mrmovq D(rB) rA 5 0 rA rB
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
rrmovq 2 0
cmovle 2 1
cmovl 2 2
cmove 2 3
cmovne 2 4
cmovge 2 5
cmovg 2 6
Y86-64 Instructions (moves)
18-600 Lecture 89262016 (copyJP Shen) 17
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
addq 6 0
subq 6 1
andq 6 2
xorq 6 3
Y86-64 Instructions (ALUs)
18-600 Lecture 89262016 (copyJP Shen) 18
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9jmp 7 0
jle 7 1
jl 7 2
je 7 3
jne 7 4
jge 7 5
jg 7 6
Y86-64 Instructions (branches)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Encoding Register Specifiers
Each register has 4-bit ID
bull Same encoding as in x86-64 Register ID 15 (0xF) indicates ldquono registerrdquo
bull Will use this in our hardware design in multiple places
rax
rcx
rdx
rbx
0
1
2
3
rsp
rbp
rsi
rdi
4
5
6
7
r8
r9
r10
r11
8
9
A
B
r12
r13
r14
No Register
C
D
E
F
9262016 (copyJP Shen) 18-600 Lecture 8 19
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Format ExampleAddition Instruction
Add value in register rA to that in register rB Store result in register rB Note that Y86-64 only allows addition to be applied to register data
Set condition codes based on result eg addq raxrsi Encoding 60 06
Two-byte encoding First indicates instruction type Second gives source and destination registers
addq rA rB 6 0 rA rB
Encoded Representation
Generic Form
9262016 (copyJP Shen) 18-600 Lecture 8 20
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Arithmetic and Logical Operations
Refer to generically as ldquoOPqrdquo Encodings differ only by ldquofunction coderdquo
Low-order 4 bits in first instruction byte
Set condition codes as side effect
addq rA rB 6 0 rA rB
subq rA rB 6 1 rA rB
andq rA rB 6 2 rA rB
xorq rA rB 6 3 rA rB
Add
Subtract (rA from rB)
And
Exclusive-Or
Instruction Code Function Code
9262016 (copyJP Shen) 18-600 Lecture 8 21
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Operations
Like the x86-64 movq instruction Simpler format for memory addresses Give different names to keep them distinct
rrmovq rA rB 2 0
Register Register
Immediate Register
irmovq V rB F rB3 0 V
Register Memory
rmmovq rA D(rB) 4 0 rA rB D
Memory Register
mrmovq D(rB) rA 5 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 22
rA rB
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
MIPS Registers$0$1$2$3$4$5$6$7$8$9$10$11$12$13$14$15
$0$at$v0$v1$a0$a1$a2$a3$t0$t1$t2$t3$t4$t5$t6$t7
Constant 0Reserved Temp
Return Values
Procedure arguments
Caller SaveTemporariesMay be overwritten bycalled procedures
$16$17$18$19$20$21$22$23$24$25$26$27$28$29$30$31
$s0$s1$s2$s3$s4$s5$s6$s7$t8$t9$k0$k1$gp$sp$s8$ra
Reserved forOperating Sys
Caller Save Temp
Global Pointer
Callee SaveTemporariesMay not beoverwritten bycalled procedures
Stack PointerCallee Save TempReturn Address
9262016 (copyJP Shen) 18-600 Lecture 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
MIPS Instruction Examples
Op Ra Rb Offset
Op Ra Rb Rd Fn00000R-R
Op Ra Rb ImmediateR-I
LoadStore
addu $3$2$1 Register add $3 = $2+$1
addu $3$2 3145 Immediate add $3 = $2+3145
sll $3$22 Shift left $3 = $2 ltlt 2
lw $316($2) Load Word $3 = M[$2+16]
sw $316($2) Store Word M[$2+16] = $3
Op Ra Rb OffsetBranch
beq $3$2dest Branch when $3 = $2
9262016 (copyJP Shen) 18-600 Lecture 8 10
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Evolutions of MIPS and ARM ISAs
9262016 (copyJP Shen) 18-600 Lecture 8 11
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ISA Design Summary
How Important is ISA Design Less now than before
With enough hardware can make almost anything go fast Installed based of legacy code and SW development tools dominate
Y86-64 Instruction Set Architecture Similar state and instructions as x86-64 Simpler encodings Somewhere between CISC and RISC
9262016 (copyJP Shen) 18-600 Lecture 8 12
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Processor Statebull Program Registers
bull 15 registers (omit r15) Each 64 bits
bull Condition Codesbull Single-bit flags set by arithmetic or logical instructions
bull ZF Zero SFNegative OF Overflow
bull Program Counterbull Indicates address of next instruction
bull Program Statusbull Indicates either normal operation or some error condition
bull Memorybull Byte-addressable storage arraybull Words stored in little-endian byte order
ZFSFOF
RF Program registers
CC Condition
codes
PC
DMEM Memory
Stat Program status
r8
r9
r10
r11
r12
r13
r14
rax
rcx
rdx
rbx
rsp
rbp
rsi
rdi
9262016 (copyJP Shen) 18-600 Lecture 8 13
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Set
18-600 Lecture 89262016 (copyJP Shen) 14
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Formats
Formatbull 1ndash10 bytes of information read from memory
bull Can determine instruction length from first bytebull Fewer instruction types and simpler encoding than x86-64
bull Each instruction accesses and modifies some part(s) of the program state
9262016 (copyJP Shen) 18-600 Lecture 8 15
18-600 Lecture 89262016 (copyJP Shen) 16
0 1 2 3 4 5 6 7 8 9
V
D
D
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB
rmmovq rA D(rB) 4 0 rA rB
mrmovq D(rB) rA 5 0 rA rB
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
rrmovq 2 0
cmovle 2 1
cmovl 2 2
cmove 2 3
cmovne 2 4
cmovge 2 5
cmovg 2 6
Y86-64 Instructions (moves)
18-600 Lecture 89262016 (copyJP Shen) 17
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
addq 6 0
subq 6 1
andq 6 2
xorq 6 3
Y86-64 Instructions (ALUs)
18-600 Lecture 89262016 (copyJP Shen) 18
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9jmp 7 0
jle 7 1
jl 7 2
je 7 3
jne 7 4
jge 7 5
jg 7 6
Y86-64 Instructions (branches)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Encoding Register Specifiers
Each register has 4-bit ID
bull Same encoding as in x86-64 Register ID 15 (0xF) indicates ldquono registerrdquo
bull Will use this in our hardware design in multiple places
rax
rcx
rdx
rbx
0
1
2
3
rsp
rbp
rsi
rdi
4
5
6
7
r8
r9
r10
r11
8
9
A
B
r12
r13
r14
No Register
C
D
E
F
9262016 (copyJP Shen) 18-600 Lecture 8 19
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Format ExampleAddition Instruction
Add value in register rA to that in register rB Store result in register rB Note that Y86-64 only allows addition to be applied to register data
Set condition codes based on result eg addq raxrsi Encoding 60 06
Two-byte encoding First indicates instruction type Second gives source and destination registers
addq rA rB 6 0 rA rB
Encoded Representation
Generic Form
9262016 (copyJP Shen) 18-600 Lecture 8 20
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Arithmetic and Logical Operations
Refer to generically as ldquoOPqrdquo Encodings differ only by ldquofunction coderdquo
Low-order 4 bits in first instruction byte
Set condition codes as side effect
addq rA rB 6 0 rA rB
subq rA rB 6 1 rA rB
andq rA rB 6 2 rA rB
xorq rA rB 6 3 rA rB
Add
Subtract (rA from rB)
And
Exclusive-Or
Instruction Code Function Code
9262016 (copyJP Shen) 18-600 Lecture 8 21
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Operations
Like the x86-64 movq instruction Simpler format for memory addresses Give different names to keep them distinct
rrmovq rA rB 2 0
Register Register
Immediate Register
irmovq V rB F rB3 0 V
Register Memory
rmmovq rA D(rB) 4 0 rA rB D
Memory Register
mrmovq D(rB) rA 5 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 22
rA rB
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
MIPS Instruction Examples
Op Ra Rb Offset
Op Ra Rb Rd Fn00000R-R
Op Ra Rb ImmediateR-I
LoadStore
addu $3$2$1 Register add $3 = $2+$1
addu $3$2 3145 Immediate add $3 = $2+3145
sll $3$22 Shift left $3 = $2 ltlt 2
lw $316($2) Load Word $3 = M[$2+16]
sw $316($2) Store Word M[$2+16] = $3
Op Ra Rb OffsetBranch
beq $3$2dest Branch when $3 = $2
9262016 (copyJP Shen) 18-600 Lecture 8 10
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Evolutions of MIPS and ARM ISAs
9262016 (copyJP Shen) 18-600 Lecture 8 11
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ISA Design Summary
How Important is ISA Design Less now than before
With enough hardware can make almost anything go fast Installed based of legacy code and SW development tools dominate
Y86-64 Instruction Set Architecture Similar state and instructions as x86-64 Simpler encodings Somewhere between CISC and RISC
9262016 (copyJP Shen) 18-600 Lecture 8 12
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Processor Statebull Program Registers
bull 15 registers (omit r15) Each 64 bits
bull Condition Codesbull Single-bit flags set by arithmetic or logical instructions
bull ZF Zero SFNegative OF Overflow
bull Program Counterbull Indicates address of next instruction
bull Program Statusbull Indicates either normal operation or some error condition
bull Memorybull Byte-addressable storage arraybull Words stored in little-endian byte order
ZFSFOF
RF Program registers
CC Condition
codes
PC
DMEM Memory
Stat Program status
r8
r9
r10
r11
r12
r13
r14
rax
rcx
rdx
rbx
rsp
rbp
rsi
rdi
9262016 (copyJP Shen) 18-600 Lecture 8 13
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Set
18-600 Lecture 89262016 (copyJP Shen) 14
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Formats
Formatbull 1ndash10 bytes of information read from memory
bull Can determine instruction length from first bytebull Fewer instruction types and simpler encoding than x86-64
bull Each instruction accesses and modifies some part(s) of the program state
9262016 (copyJP Shen) 18-600 Lecture 8 15
18-600 Lecture 89262016 (copyJP Shen) 16
0 1 2 3 4 5 6 7 8 9
V
D
D
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB
rmmovq rA D(rB) 4 0 rA rB
mrmovq D(rB) rA 5 0 rA rB
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
rrmovq 2 0
cmovle 2 1
cmovl 2 2
cmove 2 3
cmovne 2 4
cmovge 2 5
cmovg 2 6
Y86-64 Instructions (moves)
18-600 Lecture 89262016 (copyJP Shen) 17
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
addq 6 0
subq 6 1
andq 6 2
xorq 6 3
Y86-64 Instructions (ALUs)
18-600 Lecture 89262016 (copyJP Shen) 18
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9jmp 7 0
jle 7 1
jl 7 2
je 7 3
jne 7 4
jge 7 5
jg 7 6
Y86-64 Instructions (branches)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Encoding Register Specifiers
Each register has 4-bit ID
bull Same encoding as in x86-64 Register ID 15 (0xF) indicates ldquono registerrdquo
bull Will use this in our hardware design in multiple places
rax
rcx
rdx
rbx
0
1
2
3
rsp
rbp
rsi
rdi
4
5
6
7
r8
r9
r10
r11
8
9
A
B
r12
r13
r14
No Register
C
D
E
F
9262016 (copyJP Shen) 18-600 Lecture 8 19
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Format ExampleAddition Instruction
Add value in register rA to that in register rB Store result in register rB Note that Y86-64 only allows addition to be applied to register data
Set condition codes based on result eg addq raxrsi Encoding 60 06
Two-byte encoding First indicates instruction type Second gives source and destination registers
addq rA rB 6 0 rA rB
Encoded Representation
Generic Form
9262016 (copyJP Shen) 18-600 Lecture 8 20
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Arithmetic and Logical Operations
Refer to generically as ldquoOPqrdquo Encodings differ only by ldquofunction coderdquo
Low-order 4 bits in first instruction byte
Set condition codes as side effect
addq rA rB 6 0 rA rB
subq rA rB 6 1 rA rB
andq rA rB 6 2 rA rB
xorq rA rB 6 3 rA rB
Add
Subtract (rA from rB)
And
Exclusive-Or
Instruction Code Function Code
9262016 (copyJP Shen) 18-600 Lecture 8 21
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Operations
Like the x86-64 movq instruction Simpler format for memory addresses Give different names to keep them distinct
rrmovq rA rB 2 0
Register Register
Immediate Register
irmovq V rB F rB3 0 V
Register Memory
rmmovq rA D(rB) 4 0 rA rB D
Memory Register
mrmovq D(rB) rA 5 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 22
rA rB
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Evolutions of MIPS and ARM ISAs
9262016 (copyJP Shen) 18-600 Lecture 8 11
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ISA Design Summary
How Important is ISA Design Less now than before
With enough hardware can make almost anything go fast Installed based of legacy code and SW development tools dominate
Y86-64 Instruction Set Architecture Similar state and instructions as x86-64 Simpler encodings Somewhere between CISC and RISC
9262016 (copyJP Shen) 18-600 Lecture 8 12
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Processor Statebull Program Registers
bull 15 registers (omit r15) Each 64 bits
bull Condition Codesbull Single-bit flags set by arithmetic or logical instructions
bull ZF Zero SFNegative OF Overflow
bull Program Counterbull Indicates address of next instruction
bull Program Statusbull Indicates either normal operation or some error condition
bull Memorybull Byte-addressable storage arraybull Words stored in little-endian byte order
ZFSFOF
RF Program registers
CC Condition
codes
PC
DMEM Memory
Stat Program status
r8
r9
r10
r11
r12
r13
r14
rax
rcx
rdx
rbx
rsp
rbp
rsi
rdi
9262016 (copyJP Shen) 18-600 Lecture 8 13
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Set
18-600 Lecture 89262016 (copyJP Shen) 14
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Formats
Formatbull 1ndash10 bytes of information read from memory
bull Can determine instruction length from first bytebull Fewer instruction types and simpler encoding than x86-64
bull Each instruction accesses and modifies some part(s) of the program state
9262016 (copyJP Shen) 18-600 Lecture 8 15
18-600 Lecture 89262016 (copyJP Shen) 16
0 1 2 3 4 5 6 7 8 9
V
D
D
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB
rmmovq rA D(rB) 4 0 rA rB
mrmovq D(rB) rA 5 0 rA rB
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
rrmovq 2 0
cmovle 2 1
cmovl 2 2
cmove 2 3
cmovne 2 4
cmovge 2 5
cmovg 2 6
Y86-64 Instructions (moves)
18-600 Lecture 89262016 (copyJP Shen) 17
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
addq 6 0
subq 6 1
andq 6 2
xorq 6 3
Y86-64 Instructions (ALUs)
18-600 Lecture 89262016 (copyJP Shen) 18
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9jmp 7 0
jle 7 1
jl 7 2
je 7 3
jne 7 4
jge 7 5
jg 7 6
Y86-64 Instructions (branches)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Encoding Register Specifiers
Each register has 4-bit ID
bull Same encoding as in x86-64 Register ID 15 (0xF) indicates ldquono registerrdquo
bull Will use this in our hardware design in multiple places
rax
rcx
rdx
rbx
0
1
2
3
rsp
rbp
rsi
rdi
4
5
6
7
r8
r9
r10
r11
8
9
A
B
r12
r13
r14
No Register
C
D
E
F
9262016 (copyJP Shen) 18-600 Lecture 8 19
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Format ExampleAddition Instruction
Add value in register rA to that in register rB Store result in register rB Note that Y86-64 only allows addition to be applied to register data
Set condition codes based on result eg addq raxrsi Encoding 60 06
Two-byte encoding First indicates instruction type Second gives source and destination registers
addq rA rB 6 0 rA rB
Encoded Representation
Generic Form
9262016 (copyJP Shen) 18-600 Lecture 8 20
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Arithmetic and Logical Operations
Refer to generically as ldquoOPqrdquo Encodings differ only by ldquofunction coderdquo
Low-order 4 bits in first instruction byte
Set condition codes as side effect
addq rA rB 6 0 rA rB
subq rA rB 6 1 rA rB
andq rA rB 6 2 rA rB
xorq rA rB 6 3 rA rB
Add
Subtract (rA from rB)
And
Exclusive-Or
Instruction Code Function Code
9262016 (copyJP Shen) 18-600 Lecture 8 21
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Operations
Like the x86-64 movq instruction Simpler format for memory addresses Give different names to keep them distinct
rrmovq rA rB 2 0
Register Register
Immediate Register
irmovq V rB F rB3 0 V
Register Memory
rmmovq rA D(rB) 4 0 rA rB D
Memory Register
mrmovq D(rB) rA 5 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 22
rA rB
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ISA Design Summary
How Important is ISA Design Less now than before
With enough hardware can make almost anything go fast Installed based of legacy code and SW development tools dominate
Y86-64 Instruction Set Architecture Similar state and instructions as x86-64 Simpler encodings Somewhere between CISC and RISC
9262016 (copyJP Shen) 18-600 Lecture 8 12
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Processor Statebull Program Registers
bull 15 registers (omit r15) Each 64 bits
bull Condition Codesbull Single-bit flags set by arithmetic or logical instructions
bull ZF Zero SFNegative OF Overflow
bull Program Counterbull Indicates address of next instruction
bull Program Statusbull Indicates either normal operation or some error condition
bull Memorybull Byte-addressable storage arraybull Words stored in little-endian byte order
ZFSFOF
RF Program registers
CC Condition
codes
PC
DMEM Memory
Stat Program status
r8
r9
r10
r11
r12
r13
r14
rax
rcx
rdx
rbx
rsp
rbp
rsi
rdi
9262016 (copyJP Shen) 18-600 Lecture 8 13
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Set
18-600 Lecture 89262016 (copyJP Shen) 14
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Formats
Formatbull 1ndash10 bytes of information read from memory
bull Can determine instruction length from first bytebull Fewer instruction types and simpler encoding than x86-64
bull Each instruction accesses and modifies some part(s) of the program state
9262016 (copyJP Shen) 18-600 Lecture 8 15
18-600 Lecture 89262016 (copyJP Shen) 16
0 1 2 3 4 5 6 7 8 9
V
D
D
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB
rmmovq rA D(rB) 4 0 rA rB
mrmovq D(rB) rA 5 0 rA rB
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
rrmovq 2 0
cmovle 2 1
cmovl 2 2
cmove 2 3
cmovne 2 4
cmovge 2 5
cmovg 2 6
Y86-64 Instructions (moves)
18-600 Lecture 89262016 (copyJP Shen) 17
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
addq 6 0
subq 6 1
andq 6 2
xorq 6 3
Y86-64 Instructions (ALUs)
18-600 Lecture 89262016 (copyJP Shen) 18
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9jmp 7 0
jle 7 1
jl 7 2
je 7 3
jne 7 4
jge 7 5
jg 7 6
Y86-64 Instructions (branches)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Encoding Register Specifiers
Each register has 4-bit ID
bull Same encoding as in x86-64 Register ID 15 (0xF) indicates ldquono registerrdquo
bull Will use this in our hardware design in multiple places
rax
rcx
rdx
rbx
0
1
2
3
rsp
rbp
rsi
rdi
4
5
6
7
r8
r9
r10
r11
8
9
A
B
r12
r13
r14
No Register
C
D
E
F
9262016 (copyJP Shen) 18-600 Lecture 8 19
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Format ExampleAddition Instruction
Add value in register rA to that in register rB Store result in register rB Note that Y86-64 only allows addition to be applied to register data
Set condition codes based on result eg addq raxrsi Encoding 60 06
Two-byte encoding First indicates instruction type Second gives source and destination registers
addq rA rB 6 0 rA rB
Encoded Representation
Generic Form
9262016 (copyJP Shen) 18-600 Lecture 8 20
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Arithmetic and Logical Operations
Refer to generically as ldquoOPqrdquo Encodings differ only by ldquofunction coderdquo
Low-order 4 bits in first instruction byte
Set condition codes as side effect
addq rA rB 6 0 rA rB
subq rA rB 6 1 rA rB
andq rA rB 6 2 rA rB
xorq rA rB 6 3 rA rB
Add
Subtract (rA from rB)
And
Exclusive-Or
Instruction Code Function Code
9262016 (copyJP Shen) 18-600 Lecture 8 21
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Operations
Like the x86-64 movq instruction Simpler format for memory addresses Give different names to keep them distinct
rrmovq rA rB 2 0
Register Register
Immediate Register
irmovq V rB F rB3 0 V
Register Memory
rmmovq rA D(rB) 4 0 rA rB D
Memory Register
mrmovq D(rB) rA 5 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 22
rA rB
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Processor Statebull Program Registers
bull 15 registers (omit r15) Each 64 bits
bull Condition Codesbull Single-bit flags set by arithmetic or logical instructions
bull ZF Zero SFNegative OF Overflow
bull Program Counterbull Indicates address of next instruction
bull Program Statusbull Indicates either normal operation or some error condition
bull Memorybull Byte-addressable storage arraybull Words stored in little-endian byte order
ZFSFOF
RF Program registers
CC Condition
codes
PC
DMEM Memory
Stat Program status
r8
r9
r10
r11
r12
r13
r14
rax
rcx
rdx
rbx
rsp
rbp
rsi
rdi
9262016 (copyJP Shen) 18-600 Lecture 8 13
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Set
18-600 Lecture 89262016 (copyJP Shen) 14
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Formats
Formatbull 1ndash10 bytes of information read from memory
bull Can determine instruction length from first bytebull Fewer instruction types and simpler encoding than x86-64
bull Each instruction accesses and modifies some part(s) of the program state
9262016 (copyJP Shen) 18-600 Lecture 8 15
18-600 Lecture 89262016 (copyJP Shen) 16
0 1 2 3 4 5 6 7 8 9
V
D
D
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB
rmmovq rA D(rB) 4 0 rA rB
mrmovq D(rB) rA 5 0 rA rB
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
rrmovq 2 0
cmovle 2 1
cmovl 2 2
cmove 2 3
cmovne 2 4
cmovge 2 5
cmovg 2 6
Y86-64 Instructions (moves)
18-600 Lecture 89262016 (copyJP Shen) 17
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
addq 6 0
subq 6 1
andq 6 2
xorq 6 3
Y86-64 Instructions (ALUs)
18-600 Lecture 89262016 (copyJP Shen) 18
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9jmp 7 0
jle 7 1
jl 7 2
je 7 3
jne 7 4
jge 7 5
jg 7 6
Y86-64 Instructions (branches)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Encoding Register Specifiers
Each register has 4-bit ID
bull Same encoding as in x86-64 Register ID 15 (0xF) indicates ldquono registerrdquo
bull Will use this in our hardware design in multiple places
rax
rcx
rdx
rbx
0
1
2
3
rsp
rbp
rsi
rdi
4
5
6
7
r8
r9
r10
r11
8
9
A
B
r12
r13
r14
No Register
C
D
E
F
9262016 (copyJP Shen) 18-600 Lecture 8 19
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Format ExampleAddition Instruction
Add value in register rA to that in register rB Store result in register rB Note that Y86-64 only allows addition to be applied to register data
Set condition codes based on result eg addq raxrsi Encoding 60 06
Two-byte encoding First indicates instruction type Second gives source and destination registers
addq rA rB 6 0 rA rB
Encoded Representation
Generic Form
9262016 (copyJP Shen) 18-600 Lecture 8 20
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Arithmetic and Logical Operations
Refer to generically as ldquoOPqrdquo Encodings differ only by ldquofunction coderdquo
Low-order 4 bits in first instruction byte
Set condition codes as side effect
addq rA rB 6 0 rA rB
subq rA rB 6 1 rA rB
andq rA rB 6 2 rA rB
xorq rA rB 6 3 rA rB
Add
Subtract (rA from rB)
And
Exclusive-Or
Instruction Code Function Code
9262016 (copyJP Shen) 18-600 Lecture 8 21
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Operations
Like the x86-64 movq instruction Simpler format for memory addresses Give different names to keep them distinct
rrmovq rA rB 2 0
Register Register
Immediate Register
irmovq V rB F rB3 0 V
Register Memory
rmmovq rA D(rB) 4 0 rA rB D
Memory Register
mrmovq D(rB) rA 5 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 22
rA rB
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Set
18-600 Lecture 89262016 (copyJP Shen) 14
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Formats
Formatbull 1ndash10 bytes of information read from memory
bull Can determine instruction length from first bytebull Fewer instruction types and simpler encoding than x86-64
bull Each instruction accesses and modifies some part(s) of the program state
9262016 (copyJP Shen) 18-600 Lecture 8 15
18-600 Lecture 89262016 (copyJP Shen) 16
0 1 2 3 4 5 6 7 8 9
V
D
D
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB
rmmovq rA D(rB) 4 0 rA rB
mrmovq D(rB) rA 5 0 rA rB
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
rrmovq 2 0
cmovle 2 1
cmovl 2 2
cmove 2 3
cmovne 2 4
cmovge 2 5
cmovg 2 6
Y86-64 Instructions (moves)
18-600 Lecture 89262016 (copyJP Shen) 17
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
addq 6 0
subq 6 1
andq 6 2
xorq 6 3
Y86-64 Instructions (ALUs)
18-600 Lecture 89262016 (copyJP Shen) 18
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9jmp 7 0
jle 7 1
jl 7 2
je 7 3
jne 7 4
jge 7 5
jg 7 6
Y86-64 Instructions (branches)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Encoding Register Specifiers
Each register has 4-bit ID
bull Same encoding as in x86-64 Register ID 15 (0xF) indicates ldquono registerrdquo
bull Will use this in our hardware design in multiple places
rax
rcx
rdx
rbx
0
1
2
3
rsp
rbp
rsi
rdi
4
5
6
7
r8
r9
r10
r11
8
9
A
B
r12
r13
r14
No Register
C
D
E
F
9262016 (copyJP Shen) 18-600 Lecture 8 19
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Format ExampleAddition Instruction
Add value in register rA to that in register rB Store result in register rB Note that Y86-64 only allows addition to be applied to register data
Set condition codes based on result eg addq raxrsi Encoding 60 06
Two-byte encoding First indicates instruction type Second gives source and destination registers
addq rA rB 6 0 rA rB
Encoded Representation
Generic Form
9262016 (copyJP Shen) 18-600 Lecture 8 20
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Arithmetic and Logical Operations
Refer to generically as ldquoOPqrdquo Encodings differ only by ldquofunction coderdquo
Low-order 4 bits in first instruction byte
Set condition codes as side effect
addq rA rB 6 0 rA rB
subq rA rB 6 1 rA rB
andq rA rB 6 2 rA rB
xorq rA rB 6 3 rA rB
Add
Subtract (rA from rB)
And
Exclusive-Or
Instruction Code Function Code
9262016 (copyJP Shen) 18-600 Lecture 8 21
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Operations
Like the x86-64 movq instruction Simpler format for memory addresses Give different names to keep them distinct
rrmovq rA rB 2 0
Register Register
Immediate Register
irmovq V rB F rB3 0 V
Register Memory
rmmovq rA D(rB) 4 0 rA rB D
Memory Register
mrmovq D(rB) rA 5 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 22
rA rB
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Instruction Formats
Formatbull 1ndash10 bytes of information read from memory
bull Can determine instruction length from first bytebull Fewer instruction types and simpler encoding than x86-64
bull Each instruction accesses and modifies some part(s) of the program state
9262016 (copyJP Shen) 18-600 Lecture 8 15
18-600 Lecture 89262016 (copyJP Shen) 16
0 1 2 3 4 5 6 7 8 9
V
D
D
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB
rmmovq rA D(rB) 4 0 rA rB
mrmovq D(rB) rA 5 0 rA rB
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
rrmovq 2 0
cmovle 2 1
cmovl 2 2
cmove 2 3
cmovne 2 4
cmovge 2 5
cmovg 2 6
Y86-64 Instructions (moves)
18-600 Lecture 89262016 (copyJP Shen) 17
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
addq 6 0
subq 6 1
andq 6 2
xorq 6 3
Y86-64 Instructions (ALUs)
18-600 Lecture 89262016 (copyJP Shen) 18
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9jmp 7 0
jle 7 1
jl 7 2
je 7 3
jne 7 4
jge 7 5
jg 7 6
Y86-64 Instructions (branches)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Encoding Register Specifiers
Each register has 4-bit ID
bull Same encoding as in x86-64 Register ID 15 (0xF) indicates ldquono registerrdquo
bull Will use this in our hardware design in multiple places
rax
rcx
rdx
rbx
0
1
2
3
rsp
rbp
rsi
rdi
4
5
6
7
r8
r9
r10
r11
8
9
A
B
r12
r13
r14
No Register
C
D
E
F
9262016 (copyJP Shen) 18-600 Lecture 8 19
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Format ExampleAddition Instruction
Add value in register rA to that in register rB Store result in register rB Note that Y86-64 only allows addition to be applied to register data
Set condition codes based on result eg addq raxrsi Encoding 60 06
Two-byte encoding First indicates instruction type Second gives source and destination registers
addq rA rB 6 0 rA rB
Encoded Representation
Generic Form
9262016 (copyJP Shen) 18-600 Lecture 8 20
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Arithmetic and Logical Operations
Refer to generically as ldquoOPqrdquo Encodings differ only by ldquofunction coderdquo
Low-order 4 bits in first instruction byte
Set condition codes as side effect
addq rA rB 6 0 rA rB
subq rA rB 6 1 rA rB
andq rA rB 6 2 rA rB
xorq rA rB 6 3 rA rB
Add
Subtract (rA from rB)
And
Exclusive-Or
Instruction Code Function Code
9262016 (copyJP Shen) 18-600 Lecture 8 21
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Operations
Like the x86-64 movq instruction Simpler format for memory addresses Give different names to keep them distinct
rrmovq rA rB 2 0
Register Register
Immediate Register
irmovq V rB F rB3 0 V
Register Memory
rmmovq rA D(rB) 4 0 rA rB D
Memory Register
mrmovq D(rB) rA 5 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 22
rA rB
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
18-600 Lecture 89262016 (copyJP Shen) 16
0 1 2 3 4 5 6 7 8 9
V
D
D
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB
rmmovq rA D(rB) 4 0 rA rB
mrmovq D(rB) rA 5 0 rA rB
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
rrmovq 2 0
cmovle 2 1
cmovl 2 2
cmove 2 3
cmovne 2 4
cmovge 2 5
cmovg 2 6
Y86-64 Instructions (moves)
18-600 Lecture 89262016 (copyJP Shen) 17
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
addq 6 0
subq 6 1
andq 6 2
xorq 6 3
Y86-64 Instructions (ALUs)
18-600 Lecture 89262016 (copyJP Shen) 18
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9jmp 7 0
jle 7 1
jl 7 2
je 7 3
jne 7 4
jge 7 5
jg 7 6
Y86-64 Instructions (branches)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Encoding Register Specifiers
Each register has 4-bit ID
bull Same encoding as in x86-64 Register ID 15 (0xF) indicates ldquono registerrdquo
bull Will use this in our hardware design in multiple places
rax
rcx
rdx
rbx
0
1
2
3
rsp
rbp
rsi
rdi
4
5
6
7
r8
r9
r10
r11
8
9
A
B
r12
r13
r14
No Register
C
D
E
F
9262016 (copyJP Shen) 18-600 Lecture 8 19
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Format ExampleAddition Instruction
Add value in register rA to that in register rB Store result in register rB Note that Y86-64 only allows addition to be applied to register data
Set condition codes based on result eg addq raxrsi Encoding 60 06
Two-byte encoding First indicates instruction type Second gives source and destination registers
addq rA rB 6 0 rA rB
Encoded Representation
Generic Form
9262016 (copyJP Shen) 18-600 Lecture 8 20
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Arithmetic and Logical Operations
Refer to generically as ldquoOPqrdquo Encodings differ only by ldquofunction coderdquo
Low-order 4 bits in first instruction byte
Set condition codes as side effect
addq rA rB 6 0 rA rB
subq rA rB 6 1 rA rB
andq rA rB 6 2 rA rB
xorq rA rB 6 3 rA rB
Add
Subtract (rA from rB)
And
Exclusive-Or
Instruction Code Function Code
9262016 (copyJP Shen) 18-600 Lecture 8 21
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Operations
Like the x86-64 movq instruction Simpler format for memory addresses Give different names to keep them distinct
rrmovq rA rB 2 0
Register Register
Immediate Register
irmovq V rB F rB3 0 V
Register Memory
rmmovq rA D(rB) 4 0 rA rB D
Memory Register
mrmovq D(rB) rA 5 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 22
rA rB
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
18-600 Lecture 89262016 (copyJP Shen) 17
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9
addq 6 0
subq 6 1
andq 6 2
xorq 6 3
Y86-64 Instructions (ALUs)
18-600 Lecture 89262016 (copyJP Shen) 18
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9jmp 7 0
jle 7 1
jl 7 2
je 7 3
jne 7 4
jge 7 5
jg 7 6
Y86-64 Instructions (branches)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Encoding Register Specifiers
Each register has 4-bit ID
bull Same encoding as in x86-64 Register ID 15 (0xF) indicates ldquono registerrdquo
bull Will use this in our hardware design in multiple places
rax
rcx
rdx
rbx
0
1
2
3
rsp
rbp
rsi
rdi
4
5
6
7
r8
r9
r10
r11
8
9
A
B
r12
r13
r14
No Register
C
D
E
F
9262016 (copyJP Shen) 18-600 Lecture 8 19
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Format ExampleAddition Instruction
Add value in register rA to that in register rB Store result in register rB Note that Y86-64 only allows addition to be applied to register data
Set condition codes based on result eg addq raxrsi Encoding 60 06
Two-byte encoding First indicates instruction type Second gives source and destination registers
addq rA rB 6 0 rA rB
Encoded Representation
Generic Form
9262016 (copyJP Shen) 18-600 Lecture 8 20
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Arithmetic and Logical Operations
Refer to generically as ldquoOPqrdquo Encodings differ only by ldquofunction coderdquo
Low-order 4 bits in first instruction byte
Set condition codes as side effect
addq rA rB 6 0 rA rB
subq rA rB 6 1 rA rB
andq rA rB 6 2 rA rB
xorq rA rB 6 3 rA rB
Add
Subtract (rA from rB)
And
Exclusive-Or
Instruction Code Function Code
9262016 (copyJP Shen) 18-600 Lecture 8 21
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Operations
Like the x86-64 movq instruction Simpler format for memory addresses Give different names to keep them distinct
rrmovq rA rB 2 0
Register Register
Immediate Register
irmovq V rB F rB3 0 V
Register Memory
rmmovq rA D(rB) 4 0 rA rB D
Memory Register
mrmovq D(rB) rA 5 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 22
rA rB
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
18-600 Lecture 89262016 (copyJP Shen) 18
Byte
pushq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 F rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
0 1 2 3 4 5 6 7 8 9jmp 7 0
jle 7 1
jl 7 2
je 7 3
jne 7 4
jge 7 5
jg 7 6
Y86-64 Instructions (branches)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Encoding Register Specifiers
Each register has 4-bit ID
bull Same encoding as in x86-64 Register ID 15 (0xF) indicates ldquono registerrdquo
bull Will use this in our hardware design in multiple places
rax
rcx
rdx
rbx
0
1
2
3
rsp
rbp
rsi
rdi
4
5
6
7
r8
r9
r10
r11
8
9
A
B
r12
r13
r14
No Register
C
D
E
F
9262016 (copyJP Shen) 18-600 Lecture 8 19
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Format ExampleAddition Instruction
Add value in register rA to that in register rB Store result in register rB Note that Y86-64 only allows addition to be applied to register data
Set condition codes based on result eg addq raxrsi Encoding 60 06
Two-byte encoding First indicates instruction type Second gives source and destination registers
addq rA rB 6 0 rA rB
Encoded Representation
Generic Form
9262016 (copyJP Shen) 18-600 Lecture 8 20
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Arithmetic and Logical Operations
Refer to generically as ldquoOPqrdquo Encodings differ only by ldquofunction coderdquo
Low-order 4 bits in first instruction byte
Set condition codes as side effect
addq rA rB 6 0 rA rB
subq rA rB 6 1 rA rB
andq rA rB 6 2 rA rB
xorq rA rB 6 3 rA rB
Add
Subtract (rA from rB)
And
Exclusive-Or
Instruction Code Function Code
9262016 (copyJP Shen) 18-600 Lecture 8 21
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Operations
Like the x86-64 movq instruction Simpler format for memory addresses Give different names to keep them distinct
rrmovq rA rB 2 0
Register Register
Immediate Register
irmovq V rB F rB3 0 V
Register Memory
rmmovq rA D(rB) 4 0 rA rB D
Memory Register
mrmovq D(rB) rA 5 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 22
rA rB
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Encoding Register Specifiers
Each register has 4-bit ID
bull Same encoding as in x86-64 Register ID 15 (0xF) indicates ldquono registerrdquo
bull Will use this in our hardware design in multiple places
rax
rcx
rdx
rbx
0
1
2
3
rsp
rbp
rsi
rdi
4
5
6
7
r8
r9
r10
r11
8
9
A
B
r12
r13
r14
No Register
C
D
E
F
9262016 (copyJP Shen) 18-600 Lecture 8 19
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Format ExampleAddition Instruction
Add value in register rA to that in register rB Store result in register rB Note that Y86-64 only allows addition to be applied to register data
Set condition codes based on result eg addq raxrsi Encoding 60 06
Two-byte encoding First indicates instruction type Second gives source and destination registers
addq rA rB 6 0 rA rB
Encoded Representation
Generic Form
9262016 (copyJP Shen) 18-600 Lecture 8 20
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Arithmetic and Logical Operations
Refer to generically as ldquoOPqrdquo Encodings differ only by ldquofunction coderdquo
Low-order 4 bits in first instruction byte
Set condition codes as side effect
addq rA rB 6 0 rA rB
subq rA rB 6 1 rA rB
andq rA rB 6 2 rA rB
xorq rA rB 6 3 rA rB
Add
Subtract (rA from rB)
And
Exclusive-Or
Instruction Code Function Code
9262016 (copyJP Shen) 18-600 Lecture 8 21
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Operations
Like the x86-64 movq instruction Simpler format for memory addresses Give different names to keep them distinct
rrmovq rA rB 2 0
Register Register
Immediate Register
irmovq V rB F rB3 0 V
Register Memory
rmmovq rA D(rB) 4 0 rA rB D
Memory Register
mrmovq D(rB) rA 5 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 22
rA rB
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Format ExampleAddition Instruction
Add value in register rA to that in register rB Store result in register rB Note that Y86-64 only allows addition to be applied to register data
Set condition codes based on result eg addq raxrsi Encoding 60 06
Two-byte encoding First indicates instruction type Second gives source and destination registers
addq rA rB 6 0 rA rB
Encoded Representation
Generic Form
9262016 (copyJP Shen) 18-600 Lecture 8 20
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Arithmetic and Logical Operations
Refer to generically as ldquoOPqrdquo Encodings differ only by ldquofunction coderdquo
Low-order 4 bits in first instruction byte
Set condition codes as side effect
addq rA rB 6 0 rA rB
subq rA rB 6 1 rA rB
andq rA rB 6 2 rA rB
xorq rA rB 6 3 rA rB
Add
Subtract (rA from rB)
And
Exclusive-Or
Instruction Code Function Code
9262016 (copyJP Shen) 18-600 Lecture 8 21
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Operations
Like the x86-64 movq instruction Simpler format for memory addresses Give different names to keep them distinct
rrmovq rA rB 2 0
Register Register
Immediate Register
irmovq V rB F rB3 0 V
Register Memory
rmmovq rA D(rB) 4 0 rA rB D
Memory Register
mrmovq D(rB) rA 5 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 22
rA rB
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Arithmetic and Logical Operations
Refer to generically as ldquoOPqrdquo Encodings differ only by ldquofunction coderdquo
Low-order 4 bits in first instruction byte
Set condition codes as side effect
addq rA rB 6 0 rA rB
subq rA rB 6 1 rA rB
andq rA rB 6 2 rA rB
xorq rA rB 6 3 rA rB
Add
Subtract (rA from rB)
And
Exclusive-Or
Instruction Code Function Code
9262016 (copyJP Shen) 18-600 Lecture 8 21
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Operations
Like the x86-64 movq instruction Simpler format for memory addresses Give different names to keep them distinct
rrmovq rA rB 2 0
Register Register
Immediate Register
irmovq V rB F rB3 0 V
Register Memory
rmmovq rA D(rB) 4 0 rA rB D
Memory Register
mrmovq D(rB) rA 5 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 22
rA rB
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Operations
Like the x86-64 movq instruction Simpler format for memory addresses Give different names to keep them distinct
rrmovq rA rB 2 0
Register Register
Immediate Register
irmovq V rB F rB3 0 V
Register Memory
rmmovq rA D(rB) 4 0 rA rB D
Memory Register
mrmovq D(rB) rA 5 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 22
rA rB
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Move Instruction Examples
irmovq $0xabcd rdxmovq $0xabcd rdx
30 82 cd ab 00 00 00 00 00 00
X86-64 Y86-64
Encoding
rrmovq rsp rbxmovq rsp rbx
20 43
mrmovq -12(rbp)rcxmovq -12(rbp)rcx
50 15 f4 ff ff ff ff ff ff ff
rmmovq rsi0x41c(rsp)movq rsi0x41c(rsp)
40 64 1c 04 00 00 00 00 00 00
Encoding
Encoding
Encoding
9262016 (copyJP Shen) 18-600 Lecture 8 23
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Conditional Move Instructions
Refer to generically as ldquocmovXXrdquo Encodings differ only by ldquofunction coderdquo Based on values of condition codes Variants of rrmovq instruction
(Conditionally) copy value from source to destination register
rrmovq rA rB
Move Unconditionally
cmovle rA rBMove When Less or Equal
cmovl rA rB
Move When Less
cmove rA rB
Move When Equal
cmovne rA rB
Move When Not Equal
cmovge rA rB
Move When Greater or Equal
cmovg rA rB
Move When Greater
2 0 rA rB
2 1 rA rB
2 2 rA rB
2 3 rA rB
2 4 rA rB
2 5 rA rB
2 6 rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 24
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
Refer to generically as ldquojXXrdquo Encodings differ only by ldquofunction coderdquo fn Based on values of condition codes Same as x86-64 counterparts Encode full destination address
Unlike PC-relative addressing seen in x86-64
jXX Dest 7 fn
Jump (Conditionally)
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 25
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Jump Instructions
jmp Dest 7 0
Jump Unconditionally
Dest
jle Dest 7 1
Jump When Less or Equal
Dest
jl Dest 7 2
Jump When Less
Dest
je Dest 7 3
Jump When Equal
Dest
jne Dest 7 4
Jump When Not Equal
Dest
jge Dest 7 5
Jump When Greater or Equal
Dest
jg Dest 7 6
Jump When Greater
Dest
9262016 (copyJP Shen) 18-600 Lecture 8 26
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Stack
Region of memory holding program data Used in Y86-64 (and x86-64) for supporting
procedure calls Stack top indicated by rsp
Address of top stack element
Stack grows toward lower addresses Bottom element is at highest address in the stackWhen pushing must first decrement stack pointer After popping increment stack pointer
rsp
bull
bull
bull
IncreasingAddresses
Stack ldquoToprdquo
Stack ldquoBottomrdquo
9262016 (copyJP Shen) 18-600 Lecture 8 27
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stack Operations
Decrement rsp by 8 Store word from rA to memory at rsp Like x86-64
Read word from memory at rsp Save in rA Increment rsp by 8 Like x86-64
pushq rA A 0 rA F
popq rA B 0 rA F
9262016 (copyJP Shen) 18-600 Lecture 8 28
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Subroutine Call and Return
Push address of next instruction onto stack Start executing instructions at Dest Like x86-64
Pop value from stack Use as address for next instruction Like x86-64
call Dest 8 0 Dest
ret 9 0
9262016 (copyJP Shen) 18-600 Lecture 8 29
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Miscellaneous Instructions
Donrsquot do anything
Stop executing instructions x86-64 has comparable instruction but canrsquot execute it in
user mode We will use it to stop the simulator Encoding ensures that program hitting memory initialized to
zero will halt
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 30
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Status Conditions
Mnemonic CodeADR 3
Mnemonic CodeINS 4
Mnemonic CodeHLT 2
Mnemonic CodeAOK 1
Normal operation
Halt instruction encountered
Bad address (either instruction or data) encountered
Invalid instruction encountered
Desired Behavior If AOK keep going Otherwise stop program execution
9262016 (copyJP Shen) 18-600 Lecture 8 31
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Writing Y86-64 CodeTry to Use C Compiler as Much as Possible
Write code in C Compile for x86-64 with gcc ndashOg ndashS
Transliterate into Y86-64 Modern compilers make this more difficult
Coding Example Find number of elements in null-terminated list
int len1(int a[])
5043
6125
7395
0
a
rArr 3
9262016 (copyJP Shen) 18-600 Lecture 8 32
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation ExampleFirst Try
Write typical array code
Compile with gcc -Og -S
Problem Hard to do array indexing on Y86-64
Since donrsquot have scaled addressing modes
Find number of elements innull-terminated list
long len(long a[])
long lenfor (len = 0 a[len] len++)
return len
L3addq $1raxcmpq $0 (rdirax8)jne L3
9262016 (copyJP Shen) 18-600 Lecture 8 33
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 2Second Try
Write C code that mimics expected Y86-64 code
Result Compiler generates exact same code as
before Compiler converts both versions into same
intermediate formlong len2(long a)
long ip = (long) along val = (long ) iplong len = 0while (val)
ip += sizeof(long)len++val = (long ) ip
return len
9262016 (copyJP Shen) 18-600 Lecture 8 34
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Code Generation Example 3
18-600 Lecture 89262016 (copyJP Shen) 35
lenirmovq $1 r8 Constant 1irmovq $8 r9 Constant 8irmovq $0 rax len = 0mrmovq (rdi) rdx val = aandq rdx rdx Test valje Done If zero goto Done
Loopaddq r8 rax len++addq r9 rdi a++mrmovq (rdi) rdx val = aandq rdx rdx Test valjne Loop If 0 goto Loop
Doneret
Register Userdi a
rax len
rdx val
r8 1
r9 8
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Sample Program Structure 1
Program starts at address 0 Must set up stack
Where located Pointer values Make sure donrsquot overwrite code
Must initialize data
init Initialization call Mainhalt
align 8 Program dataarray
Main Main function call len
len Length function
pos 0x100 Placement of stackStack
9262016 (copyJP Shen) 18-600 Lecture 8 36
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 2
Program starts at address 0 Must set up stack Must initialize data Can use symbolic names
init Set up stack pointerirmovq Stack rsp Execute main programcall Main Terminatehalt
Array of 4 elements + terminating 0align 8
Arrayquad 0x000d000d000d000dquad 0x00c000c000c000c0quad 0x0b000b000b000b00quad 0xa000a000a000a000quad 0
9262016 (copyJP Shen) 18-600 Lecture 8 37
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Y86-64 Program Structure 3
Set up call to len Follow x86-64 procedure conventions Push array address as argument
Main irmovq arrayrdi call len(array)call lenret
9262016 (copyJP Shen) 18-600 Lecture 8 38
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Assembling Y86-64 Program
Generates ldquoobject coderdquo file lenyo Actually looks like disassembler output
unixgt yas lenys
0x054 | len0x054 30f80100000000000000 | irmovq $1 r8 Constant 10x05e 30f90800000000000000 | irmovq $8 r9 Constant 80x068 30f00000000000000000 | irmovq $0 rax len = 00x072 50270000000000000000 | mrmovq (rdi) rdx val = a0x07c 6222 | andq rdx rdx Test val0x07e 73a000000000000000 | je Done If zero goto Done0x087 | Loop0x087 6080 | addq r8 rax len++0x089 6097 | addq r9 rdi a++0x08b 50270000000000000000 | mrmovq (rdi) rdx val = a0x095 6222 | andq rdx rdx Test val0x097 748700000000000000 | jne Loop If 0 goto Loop0x0a0 | Done0x0a0 90 | ret
9262016 (copyJP Shen) 18-600 Lecture 8 39
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Simulating Y86-64 Program
Instruction set simulator Computes effect of each instruction on processor state Prints changes in state from original
unixgt yis lenyo
Stopped in 33 steps at PC = 0x13 Status HLT CC Z=1 S=0 O=0Changes to registersrax 0x0000000000000000 0x0000000000000004rsp 0x0000000000000000 0x0000000000000100rdi 0x0000000000000000 0x0000000000000038r8 0x0000000000000000 0x0000000000000001r9 0x0000000000000000 0x0000000000000008
Changes to memory0x00f0 0x0000000000000000 0x00000000000000530x00f8 0x0000000000000000 0x0000000000000013
9262016 (copyJP Shen) 18-600 Lecture 8 40
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 41
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Building BlocksCombinational Logic
Compute Boolean functions of inputs
Continuously respond to input changes
Operate on data and implement control
Storage Elements Store bits Addressable memories Non-addressable registers Loaded only as clock rises
Registerfile
A
B
W dstW
srcA
valA
srcB
valB
valW
Clock
9262016 (copyJP Shen) 18-600 Lecture 8 42
Funct
A B
ALU
Deco
de 00 11MUXSel
1001
COMP
=
PriorityEncode
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
(Cache) Memory Implementation Optionskey idx key
tag data tag data
deco
der
Associative Memory(CAM)
no indexunlimited blocks
N-Way Set-Associative Memory
k-bit index2k bull N blocks
9262016 (copyJP Shen) 18-600 Lecture 8 43
indexde
code
r
Indexed Memory
k-bit index2k blocks
Indexed Memory(Multi-Ported)(2x) k-bit index(2x) 2k blocks
index
deco
der
index
deco
der
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Hardware Control Language Very simple hardware description language Can only express limited aspects of hardware operation
Parts we want to explore and modify
Data Types bool Boolean
a b c hellip int words
A B C hellip Does not specify word size---bytes 32-bit words hellip
Statements bool a = bool-expr
int A = int-expr
9262016 (copyJP Shen) 18-600 Lecture 8 44
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
HCL Operations Classify by type of value returned
Boolean Expressions Logic Operations
a ampamp b a || b a Word Comparisons
A == B A = B A lt B A lt= B A gt= B A gt B Set Membership
A in B C D raquo Same as A == B || A == C || A == D
Word Expressions Case expressions
[ a A b B c C ] Evaluate test expressions a b c hellip in sequence Return word expression A B C hellip for first successful test
9262016 (copyJP Shen) 18-600 Lecture 8 45
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Hardware StructureState
Program counter register (PC) Condition code register (CC) Register File Memories
Access same memory space Data for readingwriting program
data Instruction for reading
instructions
Instruction Flow Read instruction at address
specified by PC Process through stages Update program counter
9262016 (copyJP Shen) 18-600 Lecture 8 46
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ StagesFetch
Read instruction from instruction memory
Decode Read program registers
Execute Compute value or address
Memory Read or write data
Write Back Write program registers
PC Update program counter
Instructionmemory
Instructionmemory
PCincrement
PCincrement
CCCCALUALU
Datamemory
Datamemory
Fetch
Decode
Execute
Memory
Write back
icode ifunrA rB
valC
Registerfile
Registerfile
A BM
E
Registerfile
Registerfile
A BM
E
PC
valP
srcA srcBdstA dstB
valA valB
aluA aluB
Cnd
valE
Addr Data
valM
PCvalE valM
newPC
9262016 (copyJP Shen) 18-600 Lecture 8 47
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Decoding
Instruction Format Instruction byte icodeifun Optional register byte rArB Optional constant word valC
5 0 rA rB D
icodeifun
rArB
valC
Optional Optional
9262016 (copyJP Shen) 18-600 Lecture 8 48
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ArithmeticLogical Operation
Fetch Read 2 bytes
Decode Read operand registers
Execute Perform operation Set condition codes
Memory Do nothing
Write back Update register
PC Update Increment PC by 2
OPq rA rB 6 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 49
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ArithLog Ops
Formulate instruction execution as sequence of simple steps
Use same general form for all instructions
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSet condition code register
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 50
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing rmmovq
Fetch Read 10 bytes
Decode Read operand registers
Execute Compute effective address
Memory Write to memory
Write back Do nothing
PC Update Increment PC by 10
rmmovq rA D(rB) 4 0 rA rB D
9262016 (copyJP Shen) 18-600 Lecture 8 51
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation rmmovq
Use ALU for address computation
rmmovq rA D(rB)icodeifun larr M1[PC]rArB larr M1[PC+1]valC larr M8[PC+2]valP larr PC+10
Fetch
Read instruction byteRead register byteRead displacement DCompute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB + valCExecute
Compute effective address
M8[valE] larr valAMemory Write value to memory Writeback
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 52
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing popq
Fetch Read 2 bytes
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read from old stack pointer
Write back Update stack pointer Write result to register
PC Update Increment PC by 2
popq rA b 0 rA 8
9262016 (copyJP Shen) 18-600 Lecture 8 53
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation popq
Use ALU to increment stack pointer Must update two registers
Popped value New stack pointer
popq rAicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rsp]valB larr R[rsp]
DecodeRead stack pointerRead stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA]Memory Read from stack R[rsp] larr valER[rA] larr valM
Writeback
Update stack pointerWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 54
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Conditional Moves
Fetch Read 2 bytes
Decode Read operand registers
Execute If cnd then set destination
register to 0xF
Memory Do nothing
Write back Update register (or not)
PC Update Increment PC by 2
cmovXX rA rB 2 fn rA rB
9262016 (copyJP Shen) 18-600 Lecture 8 55
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Cond Move
Read register rA and pass through ALU Cancel move by setting destination register to 0xF
If condition codes amp move condition indicate no move
cmovXX rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte
Compute next PCvalA larr R[rA]valB larr 0
DecodeRead operand A
valE larr valB + valAIf Cond(CCifun) rB larr 0xF
ExecutePass valA through ALU(Disable register update)
MemoryR[rB] larr valEWrite
backWrite back result
PC larr valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 56
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing Jumps
Fetch Read 9 bytes Increment PC by 9
Decode Do nothing
Execute Determine whether to take
branch based on jump condition and condition codes
Memory Do nothing
Write back Do nothing
PC Update Set PC to Dest if branch
taken or to incremented PC if not branch
jXX Dest 7 fn Dest
XX XXfall thru
XX XXtarget
Not taken
Taken
9262016 (copyJP Shen) 18-600 Lecture 8 57
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation Jumps
Compute both addresses Choose based on setting of condition codes and
branch condition
jXX Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination addressFall through address
Decode
Cnd larr Cond(CCifun)Execute
Take branchMemoryWriteback
PC larr Cnd valC valPPC update Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 58
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing call
Fetch Read 9 bytes Increment PC by 9
Decode Read stack pointer
Execute Decrement stack pointer by 8
Memory Write incremented PC to
new value of stack pointer
Write back Update stack pointer
PC Update Set PC to Dest
call Dest 8 0 Dest
XX XXreturn
XX XXtarget
9262016 (copyJP Shen) 18-600 Lecture 8 59
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation call
Use ALU to decrement stack pointer Store incremented PC
call Desticodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
Fetch
Read instruction byte
Read destination address Compute return point
valB larr R[rsp]Decode
Read stack pointervalE larr valB + ndash8
ExecuteDecrement stack pointer
M8[valE] larr valPMemory Write return value on stack R[rsp] larr valEWrite
backUpdate stack pointer
PC larr valCPC update Set PC to destination
9262016 (copyJP Shen) 18-600 Lecture 8 60
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Executing ret
Fetch Read 1 byte
Decode Read stack pointer
Execute Increment stack pointer by 8
Memory Read return address from
old stack pointer
Write back Update stack pointer
PC Update Set PC to return address
ret 9 0
XX XXreturn
9262016 (copyJP Shen) 18-600 Lecture 8 61
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Stage Computation ret
Use ALU to increment stack pointer Read return address from memory
ret
icodeifun larr M1[PC]
Fetch
Read instruction byte
valA larr R[rsp]valB larr R[rsp]
DecodeRead operand stack pointerRead operand stack pointer
valE larr valB + 8Execute
Increment stack pointer
valM larr M8[valA] Memory Read return addressR[rsp] larr valEWrite
backUpdate stack pointer
PC larr valMPC update Set PC to return address
9262016 (copyJP Shen) 18-600 Lecture 8 62
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
OPq rA rBicodeifun larr M1[PC]rArB larr M1[PC+1]
valP larr PC+2
Fetch
Read instruction byteRead register byte[Read constant word]Compute next PC
valA larr R[rA]valB larr R[rB]
DecodeRead operand ARead operand B
valE larr valB OP valASet CC
ExecutePerform ALU operationSetuse cond code reg
Memory [Memory readwrite] R[rB] larr valEWrite
backWrite back ALU result[Write back memory result]
PC larr valPPC update Update PC
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
9262016 (copyJP Shen) 18-600 Lecture 8 63
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computation Steps
All instructions follow same general pattern Differ in what gets computed on each step
call Dest
Fetch
Decode
Execute
MemoryWritebackPC update
icodeifunrArBvalCvalPvalA srcAvalB srcBvalECond codevalMdstEdstMPC
icodeifun larr M1[PC]
valC larr M8[PC+1]valP larr PC+9
valB larr R[rsp]valE larr valB + ndash8
M8[valE] larr valPR[rsp] larr valE
PC larr valC
Read instruction byte[Read register byte]Read constant wordCompute next PC[Read operand A]Read operand BPerform ALU operation[Set use cond code reg]Memory readwrite Write back ALU result[Write back memory result]Update PC
9262016 (copyJP Shen) 18-600 Lecture 8 64
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computed ValuesFetch
icode Instruction codeifun Instruction functionrA Instr Register ArB Instr Register BvalC Instruction constantvalP Incremented PC
DecodesrcA Register ID AsrcB Register ID BdstE Destination Register EdstM Destination Register MvalA Register value AvalB Register value B
Execute valE ALU result Cnd Branchmove flag
Memory valM Value from
memory
9262016 (copyJP Shen) 18-600 Lecture 8 65
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ HardwareKey
Blue boxes predesigned hardware blocks Eg memories ALU
Gray boxes control logic Describe in HCL
White ovals labels for signals
Thick lines 64-bit word values
Thin lines 4-8 bit values
Dotted lines 1-bit values
9262016 (copyJP Shen) 18-600 Lecture 8 66
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Predefined Blocks PC Register containing PC Instruction memory Read 10 bytes
(PC to PC+9) Signal invalid address
Split Divide instruction byte into icode and ifun
Align Get fields for rA rB and valCInstructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 67
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Logic
Control Logic Instr Valid Is this instruction valid icode ifun Generate no-op if invalid
address Need regids Does this instruction
have a register byte Need valC Does this instruction have
a constant word Instructionmemory
PCincrement
rBicode ifun rA
PC
valC valP
Needregids
NeedvalC
Instrvalid
AlignSplitBytes 1-9Byte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 68
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
Determine instruction codeint icode = [
imem_error INOP1 imem_icode
]
Determine instruction functionint ifun = [
imem_error FNONE1 imem_ifun
]
Instructionmemory
PC
SplitByte 0
imem_error
icode ifun
9262016 (copyJP Shen) 18-600 Lecture 8 69
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Fetch Control Logic in HCL
bool need_regids =icode in IRRMOVQ IOPQ IPUSHQ IPOPQ
IIRMOVQ IRMMOVQ IMRMOVQ
bool instr_valid = icode in INOP IHALT IRRMOVQ IIRMOVQ IRMMOVQ IMRMOVQ
IOPQ IJXX ICALL IRET IPUSHQ IPOPQ
popq rA A 0 rA F
jXX Dest 7 fn Dest
popq rA B 0 rA F
call Dest 8 0 Dest
cmovXX rA rB 2 fn rA rB
irmovq V rB 3 0 8 rB V
rmmovq rA D(rB) 4 0 rA rB D
mrmovq D(rB) rA 5 0 rA rB D
OPq rA rB 6 fn rA rB
ret 9 0
nop 1 0
halt 0 0
9262016 (copyJP Shen) 18-600 Lecture 8 70
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Decode LogicRegister File
Read ports A B Write ports E M Addresses are register IDs or
15 (0xF) (no access)
Control Logic srcA srcB read port addresses dstE dstM write port addresses
rB
dstE dstM srcA srcB
Registerfile
A BM
EdstE dstM srcA srcB
icode rA
valBvalA valEvalMCnd
Signals Cnd Indicate whether or not to
perform conditional move Computed in Execute stage
9262016 (copyJP Shen) 18-600 Lecture 8 71
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
A Source
int srcA = [icode in IRRMOVQ IRMMOVQ IOPQ IPUSHQ rAicode in IPOPQ IRET RRSP1 RNONE Dont need register
]
cmovXX rA rBvalA larr R[rA]Decode Read operand A
rmmovq rA D(rB)valA larr R[rA]Decode Read operand A
popq rAvalA larr R[rsp]Decode Read stack pointer
jXX DestDecode No operand
call Dest
valA larr R[rsp]Decode Read stack pointerret
Decode No operand
OPq rA rBvalA larr R[rA]Decode Read operand A
9262016 (copyJP Shen) 18-600 Lecture 8 72
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
E Destination
int dstE = [icode in IRRMOVQ ampamp Cnd rBicode in IIRMOVQ IOPQ rBicode in IPUSHQ IPOPQ ICALL IRET RRSP1 RNONE Dont write any register
]
None
R[rsp] larr valE Update stack pointer
None
R[rB] larr valEcmovXX rA rB
Write-back
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Write-back
Write-back
Write-back
Write-back
Write-back
Conditionally write back result
R[rsp] larr valE Update stack pointer
R[rsp] larr valE Update stack pointer
R[rB] larr valEOPq rA rB
Write-back Write back result
9262016 (copyJP Shen) 18-600 Lecture 8 73
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Execute LogicUnits
ALU Implements 4 required functions Generates condition code values
CC Register with 3 condition code
bits cond
Computes conditional jumpmove flag
Control Logic Set CC Should condition code
register be loaded ALU A Input A to ALU ALU B Input B to ALU ALU fun What function should
ALU compute
CC ALU
ALUA
ALUB
ALUfun
Cnd
icode ifun valC valBvalA
valE
SetCC
cond
9262016 (copyJP Shen) 18-600 Lecture 8 74
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU A Input
int aluA = [icode in IRRMOVQ IOPQ valAicode in IIRMOVQ IRMMOVQ IMRMOVQ valCicode in ICALL IPUSHQ -8icode in IRET IPOPQ 8 Other instructions dont need ALU
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPq rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 75
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
ALU Operation
int alufun = [icode == IOPQ ifun1 ALUADD
]
valE larr valB + ndash8 Decrement stack pointer
No operation
valE larr valB + 8 Increment stack pointer
valE larr valB + valC Compute effective address
valE larr 0 + valA Pass valA through ALUcmovXX rA rB
Execute
rmmovl rA D(rB)
popq rA
jXX Dest
call Dest
ret
Execute
Execute
Execute
Execute
Execute valE larr valB + 8 Increment stack pointer
valE larr valB OP valA Perform ALU operationOPl rA rB
Execute
9262016 (copyJP Shen) 18-600 Lecture 8 76
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory Logic
Memory Reads or writes memory
word
Control Logic stat What is instruction
status Mem read should word be
read Mem write should word be
written Mem addr Select address Mem data Select data
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
9262016 (copyJP Shen) 18-600 Lecture 8 77
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Status
Control Logic stat What is instruction
status
Datamemory
Memread
Memaddr
read
write
data out
Memdata
valE
valM
valA valP
Memwrite
data in
icode
Stat
dmem_error
instr_valid
imem_error
stat
Determine instruction statusint Stat = [
imem_error || dmem_error SADRinstr_valid SINSicode == IHALT SHLT1 SAOK
]
9262016 (copyJP Shen) 18-600 Lecture 8 78
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory AddressOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
int mem_addr = [icode in IRMMOVQ IPUSHQ ICALL IMRMOVQ valEicode in IPOPQ IRET valA Other instructions dont need address
]
9262016 (copyJP Shen) 18-600 Lecture 8 79
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Memory ReadOPq rA rB
Memory
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
No operation
M8[valE] larr valAMemory Write value to memory
valM larr M8[valA]Memory Read from stack
M8[valE] larr valPMemory Write return value on stack
valM larr M8[valA] Memory Read return address
Memory No operation
bool mem_read = icode in IMRMOVQ IPOPQ IRET
9262016 (copyJP Shen) 18-600 Lecture 8 80
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
PC Update OPq rA rB
rmmovq rA D(rB)
popq rA
jXX Dest
call Dest
ret
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr valPPC update Update PC
PC larr Cnd valC valPPC update Update PC
PC larr valCPC update Set PC to destination
PC larr valMPC update Set PC to return address
int new_pc = [icode == ICALL valCicode == IJXX ampamp Cnd valCicode == IRET valM1 valP
]
9262016 (copyJP Shen) 18-600 Lecture 8 81
New PC Select next value of PC
NewPC
Cndicode valC valPvalM
PC
Recitation
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ OperationState
PC register Cond Code register Data memory Register fileUpdated as clock rises
Combinational Logic ALU Control logic Memory reads
Instruction memory Register file Data memory
Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 82
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 2
state set according to second irmovq instruction combinational logic starting to react to state changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4Combinationallogic
Datamemory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 83
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 3
state set according to second irmovq instruction combinational logic generates results for addq instruction
Combinationallogic Data
memory
Registerfile
rbx = 0x100
PC0x014
CC100
Readports
Writeports
0x016
000
rbxlt--
0x300
Read Write
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
9262016 (copyJP Shen) 18-600 Lecture 8 84
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 4
state set according to addq instruction combinational logic starting to react to state
changes
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 85
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
SEQ Operation 5
state set according to addq instruction combinational logic generates results for je
instruction
0x014 addq rdxrbx rbx lt-- 0x300 CC lt-- 000
0x016 je dest Not taken
0x01f rmmovq rbx0(rdx) M[0x200] lt-- 0x300
Cycle 3
Cycle 4
Cycle 5
0x00a irmovq $0x200rdx rdx lt-- 0x200Cycle 2
0x000 irmovq $0x100rbx rbx lt-- 0x100Cycle 1
ClockCycle 1
Cycle 2 Cycle 3 Cycle 4
Combinationallogic
Datamemory
Registerfile
rbx = 0x300
PC0x016
CC000
Readports
Writeports
0x01f
Read Write
9262016 (copyJP Shen) 18-600 Lecture 8 86
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 8ldquoProcessor Architecture and Designrdquo
1 Processor Architecturea Instruction Set Architecture (ISA)b Instruction Set Designsc Y86-64 Instruction Set
2 Processor Implementationa Instruction Set Processor (CPU)b Processor Organization Designc Y86-64 Sequential Processor
3 Motivation for Pipelining
9262016 (copyJP Shen) 18-600 Lecture 8 87
18-600 Foundations of Computer Systems
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Computational Example
Systembull Computation requires total of 300 picosecondsbull Additional 20 picoseconds to save result in registerbull Must have clock cycle of at least 320 ps
Combinationallogic
Reg
300 ps 20 ps
Clock
Delay = 320 psThroughput = 312 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 88
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
3-Way Pipelined Version
Systembull Divide combinational logic into 3 blocks of 100 ps eachbull Can begin new operation as soon as previous one passes through stage A
bull Begin new operation every 120 psbull Overall latency increases
bull 360 ps from start to finish
Reg
Clock
Comblogic
A
Reg
Comblogic
B
Reg
Comblogic
C
100 ps 20 ps 100 ps 20 ps 100 ps 20 ps
Delay = 360 psThroughput = 833 GIPS
9262016 (copyJP Shen) 18-600 Lecture 8 89
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
16x16 combinational multiplier ISCAS-85 C6288 standard benchmark
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
[Source J Hayes Univ of Michigan]
9262016 (copyJP Shen) 18-600 Lecture 8 90
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Example Integer Multiplier
Configuration Delay MPS Area (FFwiring) Area IncreaseCombinational 352ns 284 7535 (--1759)2 Stages 187ns 534 (19x) 8725 (10781870) 164 Stages 117ns 855 (30x) 11276 (33882112) 508 Stages 080ns 1250 (44x) 17127 (89382612) 127
Pipeline efficiency 2-stage nearly double throughput marginal area cost 4-stage 75 efficiency area still reasonable 8-stage 55 efficiency area more than doubles
Tools Synopsys DCLSI Logic 110nm gflxp ASIC
9262016 (copyJP Shen) 18-600 Lecture 8 91
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining FundamentalsMotivation
bull Increase throughput with little increase in hardware
Bandwidth or Throughput = Performance
Bandwidth (BW) = no of tasksunit time For a system that operates on one task at a time
bull BW = 1delay (latency) BW can be increased by pipelining if many operands exist which need the
same operation ie many repetitions of the same task are to be performed Latency required for each task remains the same or may even increase slightly
9262016 (copyJP Shen) 18-600 Lecture 8 92
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Limitations Register Overhead
bull As we try to deepen pipeline overhead of loading registers becomes more significantbull Percentage of clock cycle spent loading register
bull 1-stage pipeline 625 bull 3-stage pipeline 1667 bull 6-stage pipeline 2857
bull High speeds of modern processor designs obtained through very deep pipelining
Delay = 420 ps Throughput = 1429 GIPSClock
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
Reg
Comblogic
50 ps 20 ps
9262016 (copyJP Shen) 18-600 Lecture 8 93
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with propagation delay T and BW = 1T
Ppipelined=BWpipelined = 1 (T k +S )
whereS = delay through latch and overhead
TS
S
Tk
Tk
k-stage pipelinedunpipelined
Pipelining Performance Model
9262016 (copyJP Shen) 18-600 Lecture 8 94
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Starting from an un-pipelined version with hardware cost G
Costpipelined = kL + G
where L = cost of adding each latch andk = number of stages
GL
L
Gk
Gk
k-stage pipelinedunpipelined
Hardware Cost Model
9262016 (copyJP Shen) 18-600 Lecture 8 95
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
CostPerformance CP = [Lk + G] [1(Tk + S)] = (Lk + G) (Tk + S)
= LT + GS + LSk + GTk
Optimal CostPerformance find min CP wrt choice of k
CostPerformance Trade-off
k
CP
[Peter M Kogge 1981]
9262016 (copyJP Shen) 18-600 Lecture 8 96
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
0
1
2
3
4
5
6
7
0 10 20 30 40 50Pipeline Depth k
x104
Cos
tPer
form
ance
Rat
io (C
P)
G=175 L=41 T=400 S=22
G=175 L=21 T=400 S=11
9262016 (copyJP Shen) 18-600 Lecture 8 97
ldquoOptimalrdquo Pipeline Depth (kopt) Examples
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining Idealistic Assumptions Uniform Sub-computations
The computation to be pipelined can be evenly partitioned into uniform-latency sub-operations
Repetition of Identical ComputationsThe same computations are to be performed repeatedly on a large number
of different inputs
Repetition of Independent Computations All the repetitions of the same computation are mutually independent ie
no data dependency and no resource conflicts
9262016 (copyJP Shen) 18-600 Lecture 8 98
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Pipeline Design Uniform Sub-computations NOT
rArr balancing pipeline stages- stage quantization to yield balanced pipe stages- minimize internal fragmentation (some waiting stages)
Identical Computations NOT rArr unifying instruction types
- coalescing instruction types into one multi-function pipe- minimize external fragmentation (some idling stages)
Independent Computations NOTrArr resolving pipeline hazards
- inter-instruction dependency detection and resolution- minimize performance loss due to pipeline stalls
9262016 (copyJP Shen) 18-600 Lecture 8 99
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 9ldquoPipelined Processor Designrdquo
John P Shen amp Zhiyi YuSeptember 28 2016
9262016 (copyJP Shen) 18-600 Lecture 8 100
18-600 Foundations of Computer Systems
Required Reading Assignmentbull Chapter 4 of CSAPP (3rd edition) by Randy Bryant amp Dave OrsquoHallaron
Recommended Reference Chapters 1 and 2 of Shen and Lipasti (SnL)
| 0 | 100 | 100 | |||
| 1 | 91152 | 80556 | |||
| 2 | 57054 | 45787 | |||
| 3 | 46289333333 | 34351333333 | |||
| 4 | 41358 | 28749 | |||
| 5 | 3876 | 2548 | |||
| 6 | 37328666667 | 23377666667 | |||
| 7 | 36564 | 21942 | |||
| 8 | 36216 | 20923 | |||
| 9 | 36145777778 | 20181777778 | |||
| 10 | 3627 | 19635 | |||
| 11 | 36535636364 | 19229636364 | |||
| 12 | 36907333333 | 18930333333 | |||
| 13 | 37360615385 | 18712615385 | |||
| 14 | 37878 | 18559 | |||
| 15 | 38446666667 | 18456666667 | |||
| 16 | 39057 | 18396 | |||
| 17 | 39701647059 | 18369647059 | |||
| 18 | 40374888889 | 18371888889 | |||
| 19 | 41072210526 | 18398210526 | |||
| 20 | 4179 | 18445 | |||
| 21 | 42525333333 | 18509333333 | |||
| 22 | 43275818182 | 18588818182 | |||
| 23 | 44039478261 | 18681478261 | |||
| 24 | 44814666667 | 18785666667 | |||
| 25 | 456 | 189 | |||
| 26 | 46394307692 | 19023307692 | |||
| 27 | 47196592593 | 19154592593 | |||
| 28 | 48006 | 19293 | |||
| 29 | 48821793103 | 19437793103 | |||
| 30 | 49643333333 | 19588333333 | |||
| 31 | 50470064516 | 19744064516 | |||
| 32 | 513015 | 199045 | |||
| 33 | 52137212121 | 20069212121 | |||
| 34 | 52976823529 | 20237823529 | |||
| 35 | 5382 | 2041 | |||
| 36 | 54666444444 | 20585444444 | |||
| 37 | 55515891892 | 20763891892 | |||
| 38 | 56368105263 | 20945105263 | |||
| 39 | 57222871795 | 21128871795 | |||
| 40 | 5808 | 21315 | |||
| 41 | 58939317073 | 21503317073 | |||
| 42 | 59800666667 | 21693666667 | |||
| 43 | 60663906977 | 21885906977 | |||
| 44 | 61528909091 | 22079909091 | |||
| 45 | 62395555556 | 22275555556 | |||
| 46 | 6326373913 | 2247273913 | |||
| 47 | 64133361702 | 22671361702 | |||
| 48 | 65004333333 | 22871333333 | |||
| 49 | 65876571429 | 23072571429 | |||
| 50 | 6675 | 23275 |
| 0 | 0 | ||
| 1 | 1 | ||
| 2 | 2 | ||
| 3 | 3 | ||
| 4 | 4 | ||
| 5 | 5 | ||
| 6 | 6 | ||
| 7 | 7 | ||
| 8 | 8 | ||
| 9 | 9 | ||
| 10 | 10 | ||
| 11 | 11 | ||
| 12 | 12 | ||
| 13 | 13 | ||
| 14 | 14 | ||
| 15 | 15 | ||
| 16 | 16 | ||
| 17 | 17 | ||
| 18 | 18 | ||
| 19 | 19 | ||
| 20 | 20 | ||
| 21 | 21 | ||
| 22 | 22 | ||
| 23 | 23 | ||
| 24 | 24 | ||
| 25 | 25 | ||
| 26 | 26 | ||
| 27 | 27 | ||
| 28 | 28 | ||
| 29 | 29 | ||
| 30 | 30 | ||
| 31 | 31 | ||
| 32 | 32 | ||
| 33 | 33 | ||
| 34 | 34 | ||
| 35 | 35 | ||
| 36 | 36 | ||
| 37 | 37 | ||
| 38 | 38 | ||
| 39 | 39 | ||
| 40 | 40 | ||
| 41 | 41 | ||
| 42 | 42 | ||
| 43 | 43 | ||
| 44 | 44 | ||
| 45 | 45 | ||
| 46 | 46 | ||
| 47 | 47 | ||
| 48 | 48 | ||
| 49 | 49 | ||
| 50 | 50 |
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining Idealistic Assumptions Uniform Sub-computations
The computation to be pipelined can be evenly partitioned into uniform-latency sub-operations
Repetition of Identical ComputationsThe same computations are to be performed repeatedly on a large number
of different inputs
Repetition of Independent Computations All the repetitions of the same computation are mutually independent ie
no data dependency and no resource conflicts
9262016 (copyJP Shen) 18-600 Lecture 8 98
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Pipeline Design Uniform Sub-computations NOT
rArr balancing pipeline stages- stage quantization to yield balanced pipe stages- minimize internal fragmentation (some waiting stages)
Identical Computations NOT rArr unifying instruction types
- coalescing instruction types into one multi-function pipe- minimize external fragmentation (some idling stages)
Independent Computations NOTrArr resolving pipeline hazards
- inter-instruction dependency detection and resolution- minimize performance loss due to pipeline stalls
9262016 (copyJP Shen) 18-600 Lecture 8 99
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 9ldquoPipelined Processor Designrdquo
John P Shen amp Zhiyi YuSeptember 28 2016
9262016 (copyJP Shen) 18-600 Lecture 8 100
18-600 Foundations of Computer Systems
Required Reading Assignmentbull Chapter 4 of CSAPP (3rd edition) by Randy Bryant amp Dave OrsquoHallaron
Recommended Reference Chapters 1 and 2 of Shen and Lipasti (SnL)
| 0 | 100 | 100 | |||
| 1 | 91152 | 80556 | |||
| 2 | 57054 | 45787 | |||
| 3 | 46289333333 | 34351333333 | |||
| 4 | 41358 | 28749 | |||
| 5 | 3876 | 2548 | |||
| 6 | 37328666667 | 23377666667 | |||
| 7 | 36564 | 21942 | |||
| 8 | 36216 | 20923 | |||
| 9 | 36145777778 | 20181777778 | |||
| 10 | 3627 | 19635 | |||
| 11 | 36535636364 | 19229636364 | |||
| 12 | 36907333333 | 18930333333 | |||
| 13 | 37360615385 | 18712615385 | |||
| 14 | 37878 | 18559 | |||
| 15 | 38446666667 | 18456666667 | |||
| 16 | 39057 | 18396 | |||
| 17 | 39701647059 | 18369647059 | |||
| 18 | 40374888889 | 18371888889 | |||
| 19 | 41072210526 | 18398210526 | |||
| 20 | 4179 | 18445 | |||
| 21 | 42525333333 | 18509333333 | |||
| 22 | 43275818182 | 18588818182 | |||
| 23 | 44039478261 | 18681478261 | |||
| 24 | 44814666667 | 18785666667 | |||
| 25 | 456 | 189 | |||
| 26 | 46394307692 | 19023307692 | |||
| 27 | 47196592593 | 19154592593 | |||
| 28 | 48006 | 19293 | |||
| 29 | 48821793103 | 19437793103 | |||
| 30 | 49643333333 | 19588333333 | |||
| 31 | 50470064516 | 19744064516 | |||
| 32 | 513015 | 199045 | |||
| 33 | 52137212121 | 20069212121 | |||
| 34 | 52976823529 | 20237823529 | |||
| 35 | 5382 | 2041 | |||
| 36 | 54666444444 | 20585444444 | |||
| 37 | 55515891892 | 20763891892 | |||
| 38 | 56368105263 | 20945105263 | |||
| 39 | 57222871795 | 21128871795 | |||
| 40 | 5808 | 21315 | |||
| 41 | 58939317073 | 21503317073 | |||
| 42 | 59800666667 | 21693666667 | |||
| 43 | 60663906977 | 21885906977 | |||
| 44 | 61528909091 | 22079909091 | |||
| 45 | 62395555556 | 22275555556 | |||
| 46 | 6326373913 | 2247273913 | |||
| 47 | 64133361702 | 22671361702 | |||
| 48 | 65004333333 | 22871333333 | |||
| 49 | 65876571429 | 23072571429 | |||
| 50 | 6675 | 23275 |
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining Idealistic Assumptions Uniform Sub-computations
The computation to be pipelined can be evenly partitioned into uniform-latency sub-operations
Repetition of Identical ComputationsThe same computations are to be performed repeatedly on a large number
of different inputs
Repetition of Independent Computations All the repetitions of the same computation are mutually independent ie
no data dependency and no resource conflicts
9262016 (copyJP Shen) 18-600 Lecture 8 98
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Pipeline Design Uniform Sub-computations NOT
rArr balancing pipeline stages- stage quantization to yield balanced pipe stages- minimize internal fragmentation (some waiting stages)
Identical Computations NOT rArr unifying instruction types
- coalescing instruction types into one multi-function pipe- minimize external fragmentation (some idling stages)
Independent Computations NOTrArr resolving pipeline hazards
- inter-instruction dependency detection and resolution- minimize performance loss due to pipeline stalls
9262016 (copyJP Shen) 18-600 Lecture 8 99
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 9ldquoPipelined Processor Designrdquo
John P Shen amp Zhiyi YuSeptember 28 2016
9262016 (copyJP Shen) 18-600 Lecture 8 100
18-600 Foundations of Computer Systems
Required Reading Assignmentbull Chapter 4 of CSAPP (3rd edition) by Randy Bryant amp Dave OrsquoHallaron
Recommended Reference Chapters 1 and 2 of Shen and Lipasti (SnL)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining Idealistic Assumptions Uniform Sub-computations
The computation to be pipelined can be evenly partitioned into uniform-latency sub-operations
Repetition of Identical ComputationsThe same computations are to be performed repeatedly on a large number
of different inputs
Repetition of Independent Computations All the repetitions of the same computation are mutually independent ie
no data dependency and no resource conflicts
9262016 (copyJP Shen) 18-600 Lecture 8 98
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Pipeline Design Uniform Sub-computations NOT
rArr balancing pipeline stages- stage quantization to yield balanced pipe stages- minimize internal fragmentation (some waiting stages)
Identical Computations NOT rArr unifying instruction types
- coalescing instruction types into one multi-function pipe- minimize external fragmentation (some idling stages)
Independent Computations NOTrArr resolving pipeline hazards
- inter-instruction dependency detection and resolution- minimize performance loss due to pipeline stalls
9262016 (copyJP Shen) 18-600 Lecture 8 99
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 9ldquoPipelined Processor Designrdquo
John P Shen amp Zhiyi YuSeptember 28 2016
9262016 (copyJP Shen) 18-600 Lecture 8 100
18-600 Foundations of Computer Systems
Required Reading Assignmentbull Chapter 4 of CSAPP (3rd edition) by Randy Bryant amp Dave OrsquoHallaron
Recommended Reference Chapters 1 and 2 of Shen and Lipasti (SnL)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining Idealistic Assumptions Uniform Sub-computations
The computation to be pipelined can be evenly partitioned into uniform-latency sub-operations
Repetition of Identical ComputationsThe same computations are to be performed repeatedly on a large number
of different inputs
Repetition of Independent Computations All the repetitions of the same computation are mutually independent ie
no data dependency and no resource conflicts
9262016 (copyJP Shen) 18-600 Lecture 8 98
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Pipeline Design Uniform Sub-computations NOT
rArr balancing pipeline stages- stage quantization to yield balanced pipe stages- minimize internal fragmentation (some waiting stages)
Identical Computations NOT rArr unifying instruction types
- coalescing instruction types into one multi-function pipe- minimize external fragmentation (some idling stages)
Independent Computations NOTrArr resolving pipeline hazards
- inter-instruction dependency detection and resolution- minimize performance loss due to pipeline stalls
9262016 (copyJP Shen) 18-600 Lecture 8 99
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 9ldquoPipelined Processor Designrdquo
John P Shen amp Zhiyi YuSeptember 28 2016
9262016 (copyJP Shen) 18-600 Lecture 8 100
18-600 Foundations of Computer Systems
Required Reading Assignmentbull Chapter 4 of CSAPP (3rd edition) by Randy Bryant amp Dave OrsquoHallaron
Recommended Reference Chapters 1 and 2 of Shen and Lipasti (SnL)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Pipelining Idealistic Assumptions Uniform Sub-computations
The computation to be pipelined can be evenly partitioned into uniform-latency sub-operations
Repetition of Identical ComputationsThe same computations are to be performed repeatedly on a large number
of different inputs
Repetition of Independent Computations All the repetitions of the same computation are mutually independent ie
no data dependency and no resource conflicts
9262016 (copyJP Shen) 18-600 Lecture 8 98
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Pipeline Design Uniform Sub-computations NOT
rArr balancing pipeline stages- stage quantization to yield balanced pipe stages- minimize internal fragmentation (some waiting stages)
Identical Computations NOT rArr unifying instruction types
- coalescing instruction types into one multi-function pipe- minimize external fragmentation (some idling stages)
Independent Computations NOTrArr resolving pipeline hazards
- inter-instruction dependency detection and resolution- minimize performance loss due to pipeline stalls
9262016 (copyJP Shen) 18-600 Lecture 8 99
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 9ldquoPipelined Processor Designrdquo
John P Shen amp Zhiyi YuSeptember 28 2016
9262016 (copyJP Shen) 18-600 Lecture 8 100
18-600 Foundations of Computer Systems
Required Reading Assignmentbull Chapter 4 of CSAPP (3rd edition) by Randy Bryant amp Dave OrsquoHallaron
Recommended Reference Chapters 1 and 2 of Shen and Lipasti (SnL)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Instruction Pipeline Design Uniform Sub-computations NOT
rArr balancing pipeline stages- stage quantization to yield balanced pipe stages- minimize internal fragmentation (some waiting stages)
Identical Computations NOT rArr unifying instruction types
- coalescing instruction types into one multi-function pipe- minimize external fragmentation (some idling stages)
Independent Computations NOTrArr resolving pipeline hazards
- inter-instruction dependency detection and resolution- minimize performance loss due to pipeline stalls
9262016 (copyJP Shen) 18-600 Lecture 8 99
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 9ldquoPipelined Processor Designrdquo
John P Shen amp Zhiyi YuSeptember 28 2016
9262016 (copyJP Shen) 18-600 Lecture 8 100
18-600 Foundations of Computer Systems
Required Reading Assignmentbull Chapter 4 of CSAPP (3rd edition) by Randy Bryant amp Dave OrsquoHallaron
Recommended Reference Chapters 1 and 2 of Shen and Lipasti (SnL)
Bryant and OrsquoHallaron Computer Systems A Programmerrsquos Perspective Third Edition
Lecture 9ldquoPipelined Processor Designrdquo
John P Shen amp Zhiyi YuSeptember 28 2016
9262016 (copyJP Shen) 18-600 Lecture 8 100
18-600 Foundations of Computer Systems
Required Reading Assignmentbull Chapter 4 of CSAPP (3rd edition) by Randy Bryant amp Dave OrsquoHallaron
Recommended Reference Chapters 1 and 2 of Shen and Lipasti (SnL)
