CSCI 136 Lab 1:135 Review

Our machine uses 32 bit memory addresses.

A memory address refers to a single byte in memory. (“Byte Addressable”)

00000000 00000000 10000000 00000000

0x00000000

0xFFFFFFFF

Memory

But our data path out of memory is designed to move 32 bit words (or multiples of 32 bits) at a time.

0x00000000

0xFFFFFFFF

Memory

Life is simpler and we get MUCH better performance if we divide memory in to 4 byte chunks (“words”) and always access these word size chunks.

Data Path

This is called “Word Alignment” 0x00000000

0xFFFFFFFF

Memory

An address is “word aligned” if its 2 least significant bits are 0… (and thus is divisible by 4)

Data Path

00

Disclaimer: You can use lb (load byte) and sb (store byte) to get a specific byte… but these commands are slow and often use lw (load word) and sw (store word) “inside the machine” along with a couple other commands to implement lb and sb.

Accessing BytesWhen might we use lb (load byte) and sb (store byte)?

Byte OrderingAssume 32 bit word: Big Endian vs. Little Endian

00000000 00000000 10000000 00000000

Byte address:0x00000000

Byte address:0x00000001

Byte address:0x00000002

Byte address:0x00000003

What is this number on a Big Endian machine?

On a Big Endian machine the Most Significant Byte (MSB) of a 32 bit word is stored at the lowest memory address.

Byte Ordering

00000000 00000000 10000000 00000000

Byte address:0x00000000

Byte address:0x00000001

Byte address:0x00000002

Byte address:0x00000003

What is the above integer on a Little Endian machine?

On a Little Endian machine the Most Significant Byte (MSB) of a 32 bit integer is stored at the highest memory address.

00000000 00000000 00000000 10000000

Byte address:0x00000000

Byte address:0x00000001

Byte address:0x00000002

Byte address:0x00000003

What is the above integer on a Little Endian machine?What is it on a Big Endian machine?

Byte OrderingComputer designers can’t seem to agree on whether to use Big Endian or Little Endian. Neither design is really superior to the other.

What kind of Endian are Intel PCs?What kind of Endian are Macs?Sun SPARC Stations can be set in either mode… (but Solaris OS requires Big Endian).

Most of the time.. you probably won’t Most of the time.. you probably won’t notice Endianness. But SPIM simulator notice Endianness. But SPIM simulator uses the “Endianness” of the machine uses the “Endianness” of the machine you run it on! you run it on!

Where does it matter?• When does “Endianness” matter?

• We will grade projects using PCSpim. If you use a Mac or XSPIM on Sun, make sure you test on a PC before submitting.

Load Immediate (li)

What does the following instruction do?

li $t0, 4

What does the following sequence of instructions do?

lui $t0, 0

ori $t0, $t0, 4

li (load immediate) instruction is really “pseudocode”

When the assembler sees an li instruction it substitutes:

lui (load upper immediate) followed by an

ori (or immediate)

In the place of the li (load immediate)

We’ll use li in the next few examples

Binary and Hex Review

• What is 0xAF (32 bit word) in decimal?

• What is 23 in binary?

2’s Complement NumbersTo get negative number from a positive..

Invert all digits and add 1.

To get a positive number from a negative… Invert all digits and add 1.

Assume 8 bit word. What is -57 in 2’s complement notation?

How do we do change a number’s sign in assembly language?

Sign Extension• Sign extension can be used when you shift

a register right… the sign bit is repeated to keep a negative number negative…

• This is referred to as “Arithmetic” variety of shift

• If the sign bit isn’t extended.. (i.e. empty spots filled with 0s) then the shift is called “Logical” variety of shift

Assume 8 bit words…What is result of logical right shifting -11 by 2?assembly: srl (shift right logical)srlv (shift right logical variable)

What is result of arithmetic right shift -11 by 2?assembly: sra (shift right arithmetic)srav (shift right arithmetic variable)

What is result of rotate right -11 by 2?assembly: ror (rotate right)

What is the assembly code to do this?

What is result of rotating left -11 by 2?assembly: rol (rotate left)

What is the result of logical shift left -11 by 2?

assembly: sll (shift left logical)

sllv (shift left logical variable)

What is the assembly code to do this?

Why isn’t there a sla (shift left arithmetic) assembly instruction?

Useful ShiftsConsidering an 8 bit word… how would we

compute dividing -6 by 2?

Consider an 8 bit word… how would we compute multiplying -7 by 4?

Useful ShiftsShifts are fast. Compilers will often

substitute shifts for multiplication and division of integers by powers of 2.

Note that rotate instructions require that rotation amount be supplied by a register. There is no “shift amount” constant value in a rotate instruction.

MIPS Adding and SubtractingWhat does following instruction do?add $t1, $t2, $t3

How about the following?addi $t1, $t2, 4

How about the following?sub $t1, $t2, $t3

What is Overflow?Consider unsigned 8 bit integers.. What

happens when we add 0x73 and 0xA9 ?

Can we get overflow when we add a positive and a negative number?

Consider signed 8 bit (2’s complement) integers.. what happens when we add 0xBC and 0xB0?

Multiplication Math Review

• What is result of multiplying unsigned integers 0xF and 0xF?

• If we multiply two 32 bit unsigned integers… what is the size in bits of the largest possible result?

Multiplication InstructionWhat does the following do?mul $t2, $t3

What do mfhi (move from hi) andmflo (move from lo) instructions do?

What does the following instruction do?

mul $t5, $t7, $t3

Memory Space

• In our SPIM simulator memory space, what percentage of our memory is reserved for the OS?

• With 32 bit addresses and byte addressing… how much memory can our machine have?

Offsets in lw and swIn our lw (load word) and sw (store word) instructions we can only supply a 16 bit offset. This offset is sign extended.

What does this command do?lw $s0, 0x0004($t0)

What does this command do?lw $s0, 0xFFFC($t0)

How many bytes can offset?

Global Pointer ($gp)Where does our data segment begin?

If we want to load the word stored at 0x10010020 the following code would be required:

lui $s0,0x1001 # What does this do?lw $s1,0x0020($s0) # and this?

This is kind of inefficient. We’re always using 2 instructions to read any word in the data segment. (example from Hennessy et al. page A-21)

Solution• Register $gp (decimal number 28) holds the

constant value 0x10008000

This allows us to offset back 32k bytes to the beginning of the data segment. Or offset forward 32k bytes.

To load the word stored at 0x10010020 we can use:lw $s1, 0x8020 ($gp)

Offsets in Branch and Jump instructions

In load word and store word instructions the 16 bit offset field of instruction counted number of bytes.number of bytes.

In branch (16 bit offset field) and jump (26 bit offset field) the offset fields count the number of instructionsnumber of instructions to offset.

Since instructions are 4 bytes.. This is also the number of words to offset…

Time Units• How long is a microsecond?

• How long is a nanosecond?

• How long is a picosecond?

• If a clock runs at 450MHZ how long is a clock cycle?

• If a clock runs at 1.2GHZ how long is clock cycle?

Cycles Per InstructionDifferent Instructions may take different numbers of

clock cycles to execute.

CPI (Cycles Per Instruction) refers to number of clock cycles an instruction takes in a design

What is average CPI of the following 600 MHZ machine? What is the MIPS rating?

Instruction Class CPI Freq

Load Word 8 31%

Store Word 7 21%

ALU (R format) 6 41%

Branch 5 5%

Jump 2 2%

• Registers $s0,$s1,…,$s7 (16 through 23) are “callee saved”.

What does this mean?

Notes on Stack Pointer:Our stack pointer will always point to

the last word in the stack. In the real world, the stack pointer occasionally points to the next free space below the stack.