chapter1 -Basic Structure of Computers · PDF filebasic structure of computers • computer types • functional units • basic operational concepts • bus structures • software

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Computer

OrganizationDesign

Chapter 1. Basic Structure of

Computers

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Basic Structure of Computers

• COMPUTER TYPES

• FUNCTIONAL UNITS

• BASIC OPERATIONAL CONCEPTS

• BUS STRUCTURES

• SOFTWARE

• PERFORMANCE

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Types of computer

Type of computer

Digital computer Analog computer Hybrid Computer

Micro ComputerMain frame

ComputerSuper Computer

Home PC

Mini Computer

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Analog computer

Analog computer measures and answer the questions by the method of “HOW MUCH”. The input data is not a number infect a physical quantity like tem, pressure, speed, velocity.

� Signals are continuous of (0 to 10 V)

�Accuracy 1% Approximately

�High speed

�Output is continuous

� Time is wasted in transmission time

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Analog computer

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Digital Computers

Digital computer counts and answer the questions by

the method of “HOW Many”. The input data is

represented by a number. These are used for the

logical and arithmetic operations.

� Signals are two level of (0 V or 5 V)

�Accuracy unlimited

� low speed sequential as well as parallel processing

�Output is continuous but obtain when computation

is completed.

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Micro Computer

Micro computer are the smallest computer system.

There size range from calculator to desktop size. Its

CPU is microprocessor. It also known as Grand child

Computer.

• Application : - personal computer, Multi user system,

offices.

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Mini Computer

These are also small general purpose system.

They are generally more powerful and most

useful as compared to micro computer. Mini

computer are also known as mid range

computer or Child computer.

• Application :- Departmental systems, Network

Servers, work group system.

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Mini computer

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Handheld

• Also called a PDA (Personal Digital Assistant).

• A computer that fits into a pocket, runs on batteries, and is used while holding in your hand.

• Typically used as an appointment book, address book, calculator, and notepad.

• Can be synchronized with a personal microcomputer as a backup.

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Handheld PC(PDA)

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Workstation

• Desktop computer which is usually more powerful than a Microcomputer.

• Powerful desktop computer designed for specialized tasks.

• A microcomputer that fits on a desk and runs on power from an electrical wall outlet.

• The CPU can be housed in either a vertical or horizontal case.

• Has separate components (keyboard, mouse, etc.) that are each plugged into the computer.

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workstation

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Laptop or Notebook

• Portable, compact computer that runs on a wall

outlet or battery unit with all components in one

unit.

• All components (keyboard, mouse, etc.) are in

one compact unit.

• Usually more expensive than a comparable

desktop.

• Sometimes called a notebook.

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Mainframe• Large powerful computer often serving many connected terminals.

• Large expensive computer capable of simultaneously processing data for hundreds or thousands of users.

• Used to store, manage, and process large amounts of data that need to be reliable, secure, and centralized.

• Usually housed in a closet sized cabinet.

• Application – Host computer, Central data base server.

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Server

• A computer that processes request for HTML

and other documents that are components of

Web pages.

• Purpose is to “serve.”

• A computer that has the purpose of supplying its

users with data; usually through the use of a

LAN (Local Area Network).

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Super Computer

• Super computer are those computer which are

designed for scientific job like whether forecasting

and artificial intelligence etc. They are fastest and

expensive. A super computer contains a number of

CPU which operate in parallel to make it faster. It also

known as grand father computer.

• Application – whether forecasting, weapons research

and development.

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Super computer

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Functional Units

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Functional Units

Figure : Basic functional units of a computer.

I/O Processor

Output

Memory

Input andArithmetic

logic

Control

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Information Handled by a Computer

• Instructions/machine instructions� Govern the transfer of information within a computer as well

as between the computer and its I/O devices

� Specify the arithmetic and logic operations to be performed

� Program

• Data� Used as operands by the instructions

� Source program

• Encoded in binary code – 0 and 1

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Memory Unit

• Store programs and data

• Two classes of storage� Primary storage

� Fast

� Programs must be stored in memory while they are being executed

� Large number of semiconductor storage cells

� Address

� RAM and memory access time

� Memory hierarchy – cache, main memory

� Secondary storage – larger and cheaper

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Arithmetic and Logic Unit (ALU)

• Most computer operations are executed in ALU of

the processor.

• Load the operands into memory – bring them to the

processor – perform operation in ALU – store the

result back to memory or retain in the processor.

• Registers

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Control Unit

• All computer operations are controlled by the control unit.

• The timing signals that govern the I/O transfers are also generated by the control unit.

• Control unit is usually distributed throughout the machine instead of standing alone.

• Operations of a computer:� Accept information in the form of programs and data through an input

unit and store it in the memory

� Fetch the information stored in the memory, under program control, into an ALU, where the information is processed

� Output the processed information through an output unit

� Control all activities inside the machine through a control unit

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Basic Operational Concepts

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Basic Operational Concepts

• Activity in a computer is governed by instructions.

• To perform a task, an appropriate program consisting of a

list of instructions is stored in the memory.

• Individual instructions are brought from the memory into

the processor, which executes the specified operations.

• Data to be used as operands are also stored in the

memory.

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A Typical Instruction

• Add LOCA, R0

– Add the operand at memory location LOCA to the operand in a register R0 in the processor.

– Place the sum into register R0.

– The original contents of LOCA are preserved.

– The original contents of R0 is overwritten.

• Instruction is fetched from the memory into the processor – the operand at LOCA is fetched and added to the contents of R0 – the resulting sum is stored in register R0.

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Separate Memory Access and ALU

Operation

• Load LOCA, R1

• Add R1, R0

• Whose contents will be overwritten?

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Connection Between the Processor

and the Memory

Figure : Connections between the processor and the memory.

Processor

Memory

PC

IR

MDR

Control

ALU

Rn 1-

R1

R0

MAR

ngeneral purposeregisters

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Registers

• Instruction register (IR)

• Program counter (PC)

• General-purpose register (R0 – Rn-1)

• Memory address register (MAR)

• Memory data register (MDR)

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Typical Operating Steps

• Programs reside in the memory through input devices

• PC is set to point to the first instruction

• The contents of PC are transferred to MAR

• A Read signal is sent to the memory

• The first instruction is read out and loaded into MDR

• The contents of MDR are transferred to IR

• Decode and execute the instruction

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Typical Operating Steps (Cont’)

• Get operands for ALU�General-purpose register

�Memory (address to MAR – Read – MDR to ALU)

• Perform operation in ALU

• Store the result back�To general-purpose register

�To memory (address to MAR, result to MDR – Write)

• During the execution, PC is incremented to the next instruction

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Operating Steps Involved for the Instruction Execution Of :

ADD LOCA, R0

• Transfer the contents of register PC to register MAR

• Issue a Read command to memory, and then wait until it has

transferred the

• requested word into register MDR

• Transfer the instruction from MDR into IR and decode it

• Transfer the address LOCA from IR to MAR

• Issue a Read command and wait until MDR is loaded

• Transfer contents of MDR to the ALU

• Transfer contents of R0 to the ALU

• Perform addition of the two operands in the ALU and transfer

result into R0

• Transfer contents of PC to ALU

• Add 1 to operand in ALU and transfer incremented address to PCvishnu 34

Interrupt

• Normal execution of programs may be preempted if

some device requires urgent servicing.

• The normal execution of the current program must be

interrupted – the device raises an interrupt signal.

• Interrupt-service routine

• Current system information backup and restore (PC,

general-purpose registers, control information, specific

information)

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Bus Structures

• There are many ways to connect different

parts inside a computer together.

• A group of lines that serves as a connecting

path for several devices is called a bus.

• Address/data/control

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Bus Structure

• Single-bus

Figure : Single-bus structure.

MemoryInput Output Processor

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Speed Issue

• Different devices have different

transfer/operate speed.

• If the speed of bus is bounded by the slowest

device connected to it, the efficiency will be

very low.

• How to solve this?

• A common approach – use buffers.

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Software

Software

– Consists of instructions that control the hardware

and cause the desired process to happen

– A Disk is considered hardware. A program ON the

disk is considered software!

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• Applications (Application Programs)

– Sets of computer instructions designed to perform

a particular application or task.

• Examples of popular application programs:

– Word or WordPerfect for word processing.

– Excel for keeping a ledger.

– Norton’s Utilities for checking disks for damage.

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• Operating Systems

– A collection of programs that manage and control all operations and coordinate all hardware components of the computer.

– Some functions include:

• Controlling the mouse pointer.

• Sending data to the printer and screen.

• Managing files.

• Formatting disks.

– Popular Operating Systems include Windows, Unix, MacOS, VMS, Linux, OS/2.

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SOFTWARE

Program

OS

routines

Hard disk

Printer

t1 t2 t3 t4t0 t5

Figure : User program and OS routines sharing the processor.

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Performance

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Performance

• The most important measure of a computer is

how quickly it can execute programs.

• Three factors affect performance:

�Hardware design

�Instruction set

�Compiler

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Performance

• Processor time to execute a program depends on the hardware involved in the execution of individual machine instructions.

Mainmemory Processor

Bus

Cachememory

Figure : The processor cache.

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Performance

• The processor and a relatively small cache

memory can be fabricated on a single

integrated circuit chip.

• Speed

• Cost

• Memory management

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Processor Clock

• Clock, clock cycle, and clock rate

• The execution of each instruction is divided

into several steps, each of which completes in

one clock cycle.

• Hertz – cycles per second

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Basic Performance Equation

• T – processor time required to execute a program that has been prepared in high-level language

• N – number of actual machine language instructions needed to complete the execution (note: loop)

• S – average number of basic steps needed to execute one machine instruction. Each step completes in one clock cycle

• R – clock rate

• Note: these are not independent to each other

R

SNT

×=

How to improve T?

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Pipeline and Superscalar Operation

• Instructions are not necessarily executed one after another.

• The value of S doesn’t have to be the number of clock cycles to execute one instruction.

• Pipelining – overlapping the execution of successive instructions.

• Superscalar operation – multiple instruction pipelines are implemented in the processor.

• Goal – reduce S (could become <1!)

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Clock Rate

• Increase clock rate� Improve the integrated-circuit (IC) technology to make the

circuits faster

� Reduce the amount of processing done in one basic step (however, this may increase the number of basic steps needed)

• Increases in R that are entirely caused by improvements in IC technology affect all aspects of the processor’s operation equally except the time to access the main memory.

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CISC and RISC

• Tradeoff between N and S

• A key consideration is the use of pipelining

�S is close to 1 even though the number of basic steps

per instruction may be considerably larger

�It is much easier to implement efficient pipelining in

processor with simple instruction sets

• Reduced Instruction Set Computers (RISC)

• Complex Instruction Set Computers (CISC)

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Compiler

• A compiler translates a high-level language program into

a sequence of machine instructions.

• To reduce N, we need a suitable machine instruction set

and a compiler that makes good use of it.

• Goal – reduce N×S

• A compiler may not be designed for a specific processor;

however, a high-quality compiler is usually designed for,

and with, a specific processor.

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Performance Measurement

• T is difficult to compute.

• Measure computer performance using benchmark programs.

• System Performance Evaluation Corporation (SPEC) selects and publishes representative application programs for different application domains, together with test results for many commercially available computers.

• Compile and run (no simulation)

• Reference computer

under testcomputer on the timeRunning

computer reference on the timeRunning =ratingSPEC

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Chapter 2. Machine

Instructions and

Programs

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Machine Instructions and Programs

� NUMBERS

� ARITHMETIC OPERATIONS AND CHARACTERS

� MEMORY LOCATIONS AND ADDRESSES

� MEMORY OPERATIONS

� INSTRUCTIONS AND INSTRUCTION SEQUENCING

� ADDRESSING MODES

vishnu 55

Number, Arithmetic

Operations, and

Characters

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Humans

• Decimal Numbers (base 10)

• Sign-Magnitude (-324)

• Decimal Fractions (23.27)

• Letters for text

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Computers• Binary Numbers (base 2)

• Two’s complement and sign-magnitude

• Binary fractions and floating point

• ASCII codes for characters (A↔65)

Why binary?

• Information is stored in computer via voltage levels.

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Decimal: Non-negatives

• Base 10

• Uses decimal digits: 0,1,2,3,4,5,6,7,8,9

• Positional System - position gives power ofthe base

• Example:3845 = 3x103 + 8x102 + 4x101 + 5x100

• Positions: …543210

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Binary: Non-negatives

• Base 2

• Uses binary digits (bits): 0,1

• Positional system

• Example:

1101 = 1x23 + 1x22 + 0x21 + 1x20

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Conversions

• External Internal

(Human) (Computer)

25 11001

A 01000001

• Humans want to see and enter numbers in

decimal.

• Computers must store and compute with bits.vishnu 61

Binary to Decimal Conversion

• Algorithm:

– Expand binary number using positional scheme.

– Perform computation using decimal arithmetic.

• Example:

110012 ↔ 1x24 + 1x23 + 0x22 + 0x21 + 1x20

= 24 + 23 + 20

= 16 + 8 + 1

= 2510

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Decimal to Binary

• Algorithm:

While N ≠ 0 doSet N to N/2Record the remainder (1 or 0)

end-of-loop

Set A to remainders in reverse order

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Decimal to binary - Example

N Rem 324

162 0

81 040 120 010 0

5 02 1

1 0

0 1

Example: Convert 32410 to binary

32410 = 1010001002

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Signed Integer

• 3 major representations:

�Sign and magnitude

�One’s complement

�Two’s complement

• Assumptions:

4-bit machine word

16 different values can be represented

Roughly half are positive, half are negative

vishnu 65

0

0

0

0

0

0

0

0

1

1

1

1

1

1

1

1

0

0

0

0

0

0

0

0

1

1

1

1

1

1

1

1

1

1

0

0

1

1

0

0

0

0

1

1

0

0

1

1

1

0

1

0

1

0

1

0

0

1

0

1

0

1

0

1 -

b3b2b1b0

B

vishnu 66

Sign and Magnitude Representation

High order bit is sign: 0 = positive (or zero), 1 = negativeThree low order bits is the magnitude: 0 (000) thru 7 (111)Number range for n bits = +/-2n-1 -1Two representations for 0

0000

0111

0011

1011

1111

1110

1101

1100

1010

1001

1000

0110

0101

0100

0010

0001

+0

+1

+2

+3

+4

+5

+6

+7-0

-1

-2

-3

-4

-5

-6

-7

0 100 = + 4

1 100 = - 4

+

-

vishnu 67

One’s Complement Representation

• Subtraction implemented by addition & 1's complement

• Still two representations of 0! This causes some problems

• Some complexities in addition

0000

0111

0011

1011

1111

1110

1101

1100

1010

1001

1000

0110

0101

0100

0010

0001

+0

+1

+2

+3

+4

+5

+6

+7-7

-6

-5

-4

-3

-2

-1

-0

0 100 = + 4

1 011 = - 4

+

-

vishnu 68

Two’s Complement Representation

• Only one representation for 0

• One more negative number than positive number

like 1's compexcept shiftedone positionclockwise

0000

0111

0011

1011

1111

1110

1101

1100

1010

1001

1000

0110

0101

0100

0010

0001

+0

+1

+2

+3

+4

+5

+6

+7-8

-7

-6

-5

-4

-3

-2

-1

0 100 = + 4

1 100 = - 4

+

-

vishnu 69

Binary, Signed-Integer

Representations

0

0

0

0

0

0

0

0

1

1

1

1

1

1

1

1

0

0

0

0

0

0

0

0

1

1

1

1

1

1

1

1

1

1

0

0

1

1

0

0

0

0

1

1

0

0

1

1

1

0

1

0

1

0

1

0

0

1

0

1

0

1

0

1

1+

1-

2+

3+

4+

5+

6+

7+

2-

3-

4-

5-

6-

7-

0+

0-

b3b2b1b0

Sign andmagnitude 1's complement 2's complement

B Values represented

Figure : Binary, signed-integer representations.vishnu 70


Representations

0

0

0

0

0

0

0

0

1

1

1

1

1

1

1

1

0

0

0

0

0

0

0

0

1

1

1

1

1

1

1

1

1

1

0

0

1

1

0

0

0

0

1

1

0

0

1

1

1

0

1

0

1

0

1

0

0

1

0

1

0

1

0

1

1+

1-

2+

3+

4+

5+

6+

7+

2-

3-

4-

5-

6-

7-

0+

0-

1+

2+

3+

4+

5+

6+

7+

0+

7-

6-

5-

4-

3-

2-

1-

0-

b3b2b1b0





Representations

0

0

0

0

0

0

0

0

1

1

1

1

1

1

1

1

0

0

0

0

0

0

0

0

1

1

1

1

1

1

1

1

1

1

0

0

1

1

0

0

0

0

1

1

0

0

1

1

1

0

1

0

1

0

1

0

0

1

0

1

0

1

0

1

1+

1-

2+

3+

4+

5+

6+

7+

2-

3-

4-

5-

6-

7-

8-

0+

0-

1+

2+

3+

4+

5+

6+

7+

0+

7-

6-

5-

4-

3-

2-

1-

0-

1+

2+

3+

4+

5+

6+

7+

0+

7-

6-

5-

4-

3-

2-

1-

b3b2b1b0




Addition of 2 Single Bits

vishnu 73

Addition

• From right to left, we add each pair of digits

• We write the sum, and add the carry to the next

column on the left

1 9 8

+ 2 6 4

Sum

Carry

0 1 1

+ 0 0 1

Sum

Carry

2

1

6

1

4

0

0

1

0

1

1

0

vishnu 74

Addition and Subtraction – 2’s Complement

4

+ 3

7

0100

0011

0111

-4

+ (-3)

-7

1100

1101

11001

4

- 3

1

0100

1101

10001

-4

+ 3

-1

1100

0011

1111

If carry-in to the high order bit =

carry-out then ignore carry

if carry-in != carry-out then overflow

Simpler addition scheme makes twos complement the most commonchoice for integer number systems within digital systems

vishnu 75

2’s-Complement Add and Subtract

Operations

1 1 0 10 1 1 1

0 1 0 0

0 0 1 01 1 0 0

1 1 1 0

0 1 1 01 1 0 1

0 0 1 1

1 0 0 10 1 0 1

1 1 1 0

1 0 0 11 1 1 1

1 0 0 0

0 0 1 00 0 1 1

0 1 0 1

4+( )

2-( )

3+( )

2-( )

8-( )

5+( )

+

+

+

+

+

+

1 1 1 0

0 1 0 01 0 1 0

0 1 1 11 1 0 1

0 1 0 0

6-( )

2-( )

4+( )

3-( )

4+( )

7+( )+

+(b)

(d)1 0 1 11 1 1 0

1 0 0 1

1 1 0 11 0 0 1

0 0 1 00 1 0 0

0 1 1 00 0 1 1

1 0 0 11 0 1 1

1 0 0 10 0 0 1

0 0 1 01 1 0 1

0 1 0 1

0 0 1 00 0 1 1

5-( )

2+( )3+( )

5+( )

2+( )4+( )

2-( )

7-( )

3-( )7-( )

6+( )3+( )

1+( )

7-( )5-( )

7-( )

2+( )3-( )

+

+

-

-

-

-

-

-

(a)

(c)

(e)

(f)

(g)

(h)

(i)

(j)

Figure : 2's-complement Add and Subtract operations.vishnu 76

Overflow - Add two positive numbers to get a negative number

or two negative numbers to get a positive number

5 + 3 = -8 -7 - 2 = +7

0000

0001

0010

0011

1000

0101

0110

0100

1001

1010

1011

1100

1101

0111

1110

1111

+0

+1

+2

+3

+4

+5

+6

+7-8

-7

-6

-5

-4

-3

-2

-1

0000

0001

0010

0011

1000

0101

0110

0100

1001

1010

1011

1100

1101

0111

1110

1111

+0

+1

+2

+3

+4

+5

+6

+7-8

-7

-6

-5

-4

-3

-2

-1

vishnu 77

Overflow Conditions

5

3

-8

0 1 1 10 1 0 1

0 0 1 1

1 0 0 0

-7

-2

7

1 0 0 01 0 0 1

1 1 0 0

1 0 1 1 1

5

2

7

0 0 0 00 1 0 1

0 0 1 0

0 1 1 1

-3

-5

-8

1 1 1 11 1 0 1

1 0 1 1

1 1 0 0 0

Overflow Overflow

No overflow No overflow

Overflow when carry-in to the high-order bit does not equal carry out

vishnu 78

Sign Extension

• Task:

– Given w-bit signed integer x

– Convert it to w+k-bit integer with same value

• Rule:

– Make k copies of sign bit:

– X ′ = xw–1

,…, xw–1

, xw–1

, xw–2

,…, x0

k copies of MSB

• • •X

X ′ • • • • • •

• • •

w

wkvishnu 79

Sign Extension Example

short int x = 15213;

int ix = (int) x;

short int y = -15213;

int iy = (int) y;

Decimal Hex Binary

x 15213 3B 6D 00111011 01101101

ix 15213 00 00 C4 92 00000000 00000000 00111011 01101101

y -15213 C4 93 11000100 10010011

iy -15213 FF FF C4 93 11111111 11111111 11000100 10010011

vishnu 80

Representation of Character

Since we are working with 0's and 1's only, to represent character in computer we use

strings of 0's and 1's only.

To represent character we are using some coding scheme, which is nothing but a mapping

function.

Some of standard coding schemes are: ASCII, EBCDIC, UNICODE.

ASCII : American Standard Code for Information Interchange.

It uses a 7-bit code. All together we have 128 combinations of 7 bits and we can represent

128 character.

As for example 65 = 1000001 represents character 'A'.

EBCDIC : Extended Binary Coded Decimal Interchange Code.

It uses 8-bit code and we can represent 256 character.

UNICODE : It is used to capture most of the languages of the world. It uses 16-bit

Unicode provides a unique number for every character, no matter what the platform, no

matter what the program, no matter what the language. The Unicode Standard has been

adopted by such industry leaders as Apple, HP, IBM, JustSystem, Microsoft, Oracle, SAP,

Sun, Sybase, Unisys and many others.vishnu 81

Memory Locations,

Addresses, and

Operations

vishnu 82

Memory Location, Addresses, and

Operation

• Memory consists of

many millions of

storage cells, each

of which can store 1

bit.

• Data is usually

accessed in n-bit

groups. n is called

word length.

• 2 k

second word

first word

Figure : Memory words.

nbits

last word

i th word

•••

•••

vishnu 83


Operation

• 32-bit word length example

(b) Four characters

charactercharactercharacter character

(a) A signed integer

Sign bit: for positive numbers

for negative numbers

ASCIIASCIIASCIIASCII

32 bits

8 bits 8 bits 8 bits 8 bits

b31 b30 b1 b0

b31 0=

b31 1=

• • •

vishnu 84


Operation

• To retrieve information from memory, either for one

word or one byte (8-bit), addresses for each location are

needed.

• A k-bit address memory has 2k memory locations, namely

0 – 2k-1, called memory space.

• 24-bit memory: 224 = 16,777,216 = 16M (1M=220)

• 32-bit memory: 232 = 4G (1G=230)

• 1K(kilo)=210

• 1T(tera)=240

vishnu 85


Operation

• It is impractical to assign distinct addresses to individual

bit locations in the memory.

• The most practical assignment is to have successive

addresses refer to successive byte locations in the

memory – byte-addressable memory.

• Byte locations have addresses 0, 1, 2, … If word length is

32 bits, the successive words are located at addresses 0,

4, 8,…

vishnu 86

Big-Endian and Little-Endian

Assignments

2k

4- 2k

3- 2k

2- 2k

1- 2k

4-2k

4-

0 1 2 3

4 5 6 7

00

4

2k

1- 2k

2- 2k

3- 2k

4-

3 2 1 0

7 6 5 4

Byte addressByte address

(a) Big-endian assignment (b) Little-endian assignment

4

Word

address

•••

•••

Figure : Byte and word addressing.

Big-Endian: lower byte addresses are used for the most significant bytes of the word

Little-Endian: opposite ordering. lower byte addresses are used for the less significant

bytes of the word

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Operation

• Address ordering of bytes

• Word alignment– Words are said to be aligned in memory if they

begin at a byte addr. that is a multiple of the numof bytes in a word.• 16-bit word: word addresses: 0, 2, 4,….

• 32-bit word: word addresses: 0, 4, 8,….

• 64-bit word: word addresses: 0, 8,16,….

• Access numbers, characters, and character strings

vishnu 88

Memory Operation

• Load (or Read or Fetch)

� Copy the content. The memory content doesn’t change.

� Address – Load

� Registers can be used

• Store (or Write)

� Overwrite the content in memory

� Address and Data – Store

� Registers can be used

vishnu 89

Instruction and

Instruction Sequencing

vishnu 90

Types of Instructions

Types of Instructions:

• Data transfers between the memory and the

processor registers

• Arithmetic and logic operations on data

• Program sequencing and control

• I/O transfers

vishnu 91

Representation of Machine Instruction

2 ways:

1. REGISTER TRANSFER NOTATION

2. ASSEMBLY LANGAGE NOTATION

vishnu 92

Register Transfer Notation

• Identify a location by a symbolic name standing for its hardware binary address (LOC, R0,…)

• Example:– Momery locations – LOC, PLACE, A,VAR2

– Processor register – R0 . . . Rn-1

– IO register - DATAIN, DATAOUT

• Contents of a location are denoted by placing square brackets around the name of the location

• Example:

[LOC]

vishnu 93

• R1←[LOC]

• R3 ←[R1]+[R2]

vishnu 94

Assembly Language Notation

• Represent machine instructions and programs.

• Instruction having the format:

• Example:

• Move LOC, R1

• Add R1, R2, R3

Opcode Operand(s) or Address(es)

vishnu 95

Basic Instruction Types

Let A, B and C are variables then

C=A+B

In register transfer notation it can be written as:

C�[A]+[B]

Types of Instructions:• Three-Address Instructions

• Two-Address Instructions

• One-Address Instructions

• Zero-Address Instructions

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Three-Address Instructions

Operation Source 1 ,Source2,Destination

Example:

ADD R1, R2, R3 R3 ← R1 + R2

For C=A+B

ADD A, B, C

What is the problem?.

vishnu 97

Two-Address Instructions

Operation Source, Destination

Example:ADD R1, R2 R2 ← R1 + R2

For C=A+B:

ADD A, B

MOV B, C

What is the problem?.

vishnu 98

One-Address Instructions

• Instruction can have only one memory operand

• Implicitly Accumulator is taken as another operand

• Example:

ADD A AC�[AC]+[A]

For C=A+B

LOAD A

ADD B

STORE C

vishnu 99

Zero-Address Instructions

• Processor uses pushdown stack

Example:ADD TOS ← TOS + (TOS – 1)

For C=A+B

PUSH A

PUSH B

ADD

POP C

vishnu 100

Summery: Basic Instruction Types

• Three-Address Instructions

– ADD R1, R2, R3 R3 ← R1 + R2


– ADD R1, R2 R2 ← R1 + R2


– ADD M AC ← AC + M[AR]


– ADD TOS ← TOS + (TOS – 1)

• RISC Instructions

– Lots of registers. Memory is restricted to Load & Store

Opcode Operand(s) or Address(es)vishnu 101

Instruction Execution and Straight-Line Sequencing

vishnu 102

• C=A+B is a High Level Language statement where

A, B and C are memory locations

• Assembly language program for C=A+B is:

MOV A, R0

ADD B, R0

MOV R0, C

vishnu 103

Instruction Execution and Straight-

Line Sequencing

R0,C

B,R0

A,R0

Movei + 8

Begin execution here Movei

ContentsAddress

C

B

A

the programData for

segmentprogram3-instruction

Addi + 4

Figure : A program for C ← [Α] + [Β].

Assumptions:

- One memory operand

per instruction

- 32-bit word length

- Memory is byte

addressable

- Full memory address

can be directly specified

in a single-word instruction

Two-phase procedure

-Instruction fetch

-Instruction execute

vishnu 104

Branching

NUMn

NUM2

NUM1

R0,SUM

NUMn,R0

NUM3,R0

NUM2,R0

NUM1,R0

Figure : A straight-line program for adding n numbers.

Add

Add

Move

SUM

i

Move

Add

i 4n+

i 4n 4-+

i 8+

i 4+

•••

•••

•••

vishnu 105

BranchingN,R1Move

NUMn

NUM2

NUM1

R0,SUM

R1

"Next" number to R0

Figure : Using a loop to add n numbers.

LOOP

Decrement

Move

LOOP

loop

Program

Determine address of"Next" number and add

N

SUM

n

R0Clear

Branch>0

•••

•••

Branch target

Conditional branch

vishnu 106

Condition Codes

• Condition code flags

• Condition code register / status register

• N (negative)

• Z (zero)

• V (overflow)

• C (carry)

• Different instructions affect different flags

vishnu 107

Summery: Basic Instruction Types

• Three-Address Instructions

– ADD R1, R2, R3 R3 ← R1 + R2


– ADD R1, R2 R2 ← R1 + R2


– ADD M AC ← AC + M[AR]


– ADD TOS ← TOS + (TOS – 1)

• RISC Instructions

– Lots of registers. Memory is restricted to Load & Store

Opcode Operand(s) or Address(es)vishnu 108

Conditional Branch Instructions

• Example:

– A: 1 1 1 1 0 0 0 0

– B: 0 0 0 1 0 1 0 0

A: 1 1 1 1 0 0 0 0

+(−B): 1 1 1 0 1 1 0 0

1 1 0 1 1 1 0 0

C = 1

S = 1

V = 0

Z = 0

vishnu 109

Addressing Modes

vishnu 110

• Register mode

– Operand is in the processor register

– Name of the register is given in the instruction

• Absolute/Direct mode

– Operand is in memory location.

– Address of the memory location is given in the

instruction

vishnu 111

• Immediate

– Operand is given explicitly in the instruction

– The use of a constant in “MOV #5, R1,”, i.e. R1 ← 5

vishnu 112

• Indirect Address

– Indicate the memory location that holds the

address of the memory location that holds the

dataAR = 101

100

101

102

103

104

0 1 0 4

1 1 0 A

i.e., Operand is avialable in memory,

That Memory address is avialable in either

register or Memory.

vishnu 113

• Index mode – the effective address of the operand is

generated by adding a constant value to the contents of a

register.

• X(Ri):

EA = X + [Ri]

• Several variations:

(Ri, Rj): EA = [Ri] + [Rj]

X(Ri, Rj): EA = X + [Ri] + [Rj]

vishnu 114

• Indexed

– EA = Index Register + Relative Addr

100

101

102

103

104

Rj = 100

1 1 0 A

Ri = 2

+

Could be Positive or Negative

(2’s Complement)

Useful with “Autoincrement” or “Autodecrement”

vishnu 115

• Relative mode – the effective address is determined by

the Index mode using the program counter in place of

the general-purpose register.

• X(PC) – note that X is a signed number

• Branch>0 LOOP

• This location is computed by specifying it as an offset

from the current value of PC.

NOTE: Branch target may be either before or after the

branch instruction, the offset is given as a singed num.

vishnu 116

100

101

102

103

104

0

1

2

• Relative Address

– EA = PC + Relative Addr

AR = 100

1 1 0 A

PC = 2

+

Could be Positive or Negative

(2’s Complement)

vishnu 117

• Autoincrement mode – the effective address of the operand is the contents of a register specified in the instruction. After accessing the operand, the contents of this register are automatically incremented to point to the next item in a list.

• (Ri)+. The increment is 1 for byte-sized operands, 2 for 16-bit operands, and 4 for 32-bit operands.

• Autodecrement mode: -(Ri) – decrement first

R0Clear

R0,SUM

R1

(R2)+,R0

Figure : The Autoincrement addressing mode used in the program

Initialization

Move

LOOP Add

Decrement

LOOP

#NUM1,R2

N,R1Move

Move

Branch>0

vishnu 118

Addressing Modes

• The different ways in which the location of an operand is specified in an instruction are referred to as addressing modes.

Name Assembler syntax Addressing function

Immediate #Value Operand = Value

Register R i EA = Ri

Absolute(Direct) LOC EA = LOC

Indirect (Ri ) EA = [Ri ]

(LOC) EA = [LOC]

Index X(R i) EA = [Ri ] + X

Basewith index (Ri ,Rj ) EA = [Ri ] + [Rj ]

Basewith index X(R i ,Rj ) EA = [Ri ] + [Rj ] + X

and offset

Relative X(PC) EA = [PC] + X

Autoincrement (Ri )+ EA = [Ri ] ;Increment R i

Autodecrement (R i ) Decrement R i ;

EA = [Ri ]

−

vishnu 119

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chapter1 -Basic Structure of Computers · PDF filebasic structure of computers • computer types • functional units • basic operational concepts • bus structures • software