Introduction To Computer Architecture Jean-Michel RICHER University of Angers France...

IntroductionTo

Computer Architecture

Jean-Michel RICHER

University of AngersFrance

[email protected]

January 2003

mailto:[email protected]

WARNING !

This document can be reproduced, modified and used freely.

Please report any change or improvement to Jean-Michel Richer at :

[email protected](http://www.info.univ-angers.fr/pub/richer)

January 2003



http://www.info.univ-angers.fr/pub/richer

OUTLINE

Memory organization

Some CPU organizations

Example of program execution with the Intel 8086

Example of program execution with pipeline technology

MEMORY ORGANIZATION

The operating system organizes the memory as follows :

contains the program codeCode

contains the program dataData

contains subprograms callsSTACK

contains the rest of the memoryHEAP

DIFFERENT CPU ORGANIZATIONS

808680486

Pentium IIIPentium IV

Code

Data

STACK

HEAP

Load Instruction

Decode

Load Operand

Execute

ALU

Registe

rs

Write Result

Intel 8086

Instruction path

Data path

Cpu diemotherboard

Coprocessor

Code

Data

STACK

HEAP

Load Instruction

Decode

Load Operand

Execute

ALU

Registe

rs

Write Result

Intel 80486

L1 Cache - I

L1 Cache

FPU

Integrated FPUL2

Cach

e

L2 Cache onMotherboard

Code

Data

STACK

HEAP

L1 Cache - I

Load Instruction

Decode

Load Operand

Execute

FPU

Registe

rs

Write Result

Intel Pentium

BTB TLB

Branch Prediction

ALU1

ALU2

Superscalar

L2 C

ach

e

L1 C

ach

e - D

L1 Cache

Code

Data

STACK

HEAP

L1 Cache - I

Load Instruction

Decode

Load Operand

Execute

ALU1

ALU2

FPU

BTB TLB

L2 C

ach

e

Registe

rs

L1 C

ach

e - D

Write Result

Intel Pentium III

L2 Cache on-die

Code

Data

STACK

HEAP

Load Instruction

Decode

Load Operand

Execute

UAL1

UAL2

FPU

BTB TLB

L2 C

ach

e

Registe

rs

L1 C

ach

e - D

Write Result

Intel Pentium IV

Trace Cache

Trace Cache

PROGRAM EXECUTION

The 8086 case

int a, b ,c

a = 1;b = 2;c = a + b

C Language

(a) 100 0000000000000001(b) 102 0000000000000010(c) 104 0000000000000000

Program

mov ax, [100]mov bx, [102]add ax, bxmov [104], ax

Assembler 8086

Micro code

OP Value 1 Value 2 R1 R2

2 bits 2 bits 2 bits16 bits 16 bits

00 mov r1,[adr]01 mov [adr],r111 add r1,r2

00 AX01 BX10 CX11 DX

Instructions are converted into micro-operations that can be handled by the CPU (Central Processing Unit) of the micro-processor

Operation Values Registers

mov ax,[100]mov bx,[102]add ax, bxmov [104],ax

[100]=1[102]=2[104]=0

STACK

HEAP

RegistersLoad Instruction

Decode

Load Operand

Execute

UAL

Write Result

AX=?BX=?CX=?DX=?

Starting point


[100]=1[102]=2[104]=0

STACK

HEAP

Registersmov ax,[100]

Decode

Load Operand

Execute

UAL

Write Result

AX=?BX=?CX=?DX=?

Load Instruction : mov ax,[100]


[100]=1[102]=2[104]=0

STACK

HEAP


00 100 ? 00 ?

Load Operand

Execute

UAL

Write Result

AX=?BX=?CX=?DX=?

Decode Instruction : mov ax,[100]

Micro code

00 00100 ??

op val1 val2 r1 r2


[100]=1[102]=2[104]=0

STACK

HEAP


00 100 ? 00 ?

00 100 1 00 ?

Execute

UAL

Write Result

AX=?BX=?CX=?DX=?

Load operand : mov ax,[100]

Micro code

00 00100 ?1

op val1 val2 r1 r2


[100]=1[102]=2[104]=0

STACK

HEAP


00 100 ? 00 ?

00 100 1 00 ?

Execute

UAL

Write Result

AX=?BX=?CX=?DX=?

Execute : mov ax,[100]


[100]=1[102]=2[104]=0

STACK

HEAP


00 100 ? 00 ?

00 100 1 00 ?

Execute

UAL

00 100 1 00 ?

AX=1BX=?CX=?DX=?

Write result : mov ax,[100]


[100]=1[102]=2[104]=0

STACK

HEAP

Registersmov bx,[102]

Decode

Load Operand

Execute

UAL

Write Result

AX=1BX=?CX=?DX=?

Load Instruction : mov bx,[102]


[100]=1[102]=2[104]=0

STACK

HEAP


00 102 ? 01 ?

Load Operand

Execute

UAL

Write Result

AX=1BX=?CX=?DX=?

Decode Instruction : mov bx,[102]

Micro code

00 01102 ??

op val1 val2 r1 r2


[100]=1[102]=2[104]=0

STACK

HEAP


00 102 ? 01 ?

00 102 2 00 ?

Execute

UAL

Write Result

AX=1BX=?CX=?DX=?

Load operand : mov bx,[102]

Micro code

00 01102 ?2

op val1 val2 r1 r2


[100]=1[102]=2[104]=0

STACK

HEAP


00 102 ? 01 ?

00 102 2 00 ?

Execute

UAL

Write Result

AX=1BX=?CX=?DX=?

Execute : mov bx,[102]


[100]=1[102]=2[104]=0

STACK

HEAP


00 102 ? 01 ?

00 102 2 01 ?

Execute

UAL

00 102 2 01 ?

AX=1BX=2CX=?DX=?

Write result : mov bx,[102]


[100]=1[102]=2[104]=0

STACK

HEAP

Registersadd ax,bx

Decode

Load Operand

Execute

UAL

Write Result

AX=1BX=2CX=?DX=?

Load Instruction : add ax,bx


[100]=1[102]=2[104]=0

STACK

HEAP

Registersadd ax,bx

11 ? ? 00 01

Load Operand

Execute

UAL

Write Result

AX=1BX=2CX=?DX=?

Decode Instruction : add ax,bx

Micro code

11 00? 01?

op val1 val2 r1 r2


[100]=1[102]=2[104]=0

STACK

HEAP

Registersadd ax,bx

11 ? ? 00 01

11 1 2 00 01

Execute

UAL

Write Result

AX=1BX=2CX=?DX=?

Load Operand : add ax,bx

Micro code

11 001 012

op val1 val2 r1 r2


[100]=1[102]=2[104]=0

STACK

HEAP

Registersadd ax,bx

11 ? ? 00 01

11 1 2 00 01

Execute

1+2 11 3 2 00 01

Write Result

AX=1BX=2CX=?DX=?

Execute : add ax,bx

Micro code

11 003 012

op val1 val2 r1 r2


[100]=1[102]=2[104]=0

STACK

HEAP

Registersadd ax,bx

11 ? ? 00 01

11 1 2 00 01

Execute

11 3 2 00 01

11 3 2 00 01

AX=3BX=2CX=?DX=?

Write Result : add ax,bx


[100]=1[102]=2[104]=0

STACK

HEAP

Registersmov [104],ax

Decode

Load Operand

Execute

UAL

Write Result

AX=3BX=2CX=?DX=?

Load Instruction : mov [104],ax


[100]=1[102]=2[104]=0

STACK

HEAP


01 104 ? 00 ??

Load Operand

Execute

UAL

Write Result

AX=3BX=2CX=?DX=?

Decode Instruction : mov [104],ax

Micro code

01 00104 ??

op val1 val2 r1 r2


[100]=1[102]=2[104]=0

STACK

HEAP


01 104 ? 00 ??

01 104 3 00 ??

Execute

UAL

Write Result

AX=3BX=2CX=?DX=?

Load Operand : mov [104],ax

Micro code

01 00104 ?3

op val1 val2 r1 r2


[100]=1[102]=2[104]=0

STACK

HEAP


01 104 ? 00 01

01 104 3 00 01

Execute

UAL

Write Result

AX=3BX=2CX=?DX=?


Micro code

01 00104 ?3

op val1 val2 r1 r2


[100]=1[102]=2[104]=3

STACK

HEAP


01 104 ? 00 01

01 104 3 00 01

Execute

UAL

01 104 3 00 01

AX=3BX=2CX=?DX=?

Write Result : mov [104],ax

PROGRAM EXECUTION WITH PIPELINE

Pentium-Like

DEPENDENCIES

mov ax, [100]

mov bx, [102]

add ax, bx

mov [104], ax

The 8086 code shows there are 2 dependencies :

RAW

WAW

Forward

Forward

AVOID DEPENDENCIES

To avoid dependencies we can add a field to the micro-code in order to indicate when data need to be forwarded into the pipeline :

OP Value 1 Value 2 R1 R2

2 bits 2 bits 2 bits16 bits 16 bits

Operation Values Registers

F

1 bit

If F = 1 then Forward Data

Flag

00 00100 ??

op val1 val2 r1 r2

TRADUCTION INTO MICRO-OPs

mov ax, [100]

mov bx, [102]

add ax, bx

mov [104], ax

00 01102 ??

11 00? 01?

01 00104 ??

0

f

1

1

0

Load Operand

Execute

UAL

Write Result

Load Operand : mov ax,[100]

00 00100 ?? 01

Step 1

Load Operand

Execute

UAL

Write Result

Load Operand : mov bx,[102]Execute : mov ax,[100]

00 01102 ?? 1

00 00100 ?1 0

2

Step 2

Load Operand

Execute

UAL

Write Result

Execute : mov bx,[102]

AX=1

Load Operand : add ax,bx

11 00? 01? 1

00 01102 ?2 1

Write Result : mov ax,[100]

00 00100 ?1 0

1

Step 3

Forward result

1

2

Load Operand

Execute

UAL

Write Result BX=2


01 00104 ?? 0

Execute : add ax, bx

11 001 012 1

Write Result : mov bx,[102]

00 01102 ?2 1

1+2

11 00 012 13

Step 4

Forward result

1

3

Load Operand

Execute

UAL

Write Result AX=3

Execute : mov ax,[104]

01 00104 ?3 0

Write Result : add ax, bx

11 003 012 1

Step 5

Load Operand

Execute

UAL

Write Result

Write Result : mov ax, [104]

01 00104 ?3 0

RAM

Step 6

Introduction To Computer Architecture Jean-Michel RICHER University of Angers France...

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