CS2422 Basic Concepts

Department of Computer ScienceNational Tsing Hua University

Assembly Language for Intel-

Based Computers, 5th Edition

Chapter 1: Basic Concepts

(c) Pearson Education, 2006-2007. All rights reserved. You may modify and copy this slide show for your personal use, or for use in the classroom, as long as this copyright statement, the author's name, and the title are not changed.

Slides prepared by the author, with modification

Revision date: June 3, 2006

Kip Irvine

當你在寫 C 程式時，心中想像的電腦是長什麼樣子？

CPU, memory, I/O,Operations on variables,

…

4

A Model of Computer for C

+

if for

&&

CPU

Memory

i, j, k;

xfloat, yfloat;

A[0], A[1], …

i = i + j;xfloat = 1.0;if (A[0]==0)…

ProgramCounter

Typed storage

這個model和你實際跑程式的電腦有很大的差別！

為什麼你用這個 C 的 model所寫的程式能在你的電腦上執行？

顯然有人為你做了轉換！

6

MOV AX, aADD AX, bMUL cMOV x, AX

MOV AX, aADD AX, bMUL cMOV x, AX

x = (a+b) * bx = (a+b) * b

我們已經知道 ...

C compilerC compiler

Assembly program

C program

可見得組合語言程式更靠近實際的電腦

組合語言程看到的電腦是長什麼樣子 ?

8

A Model of Computer for ASM

CPU

MemoryMOV AX, aADD AX, bMOV x, AX…

010100110010101a

110010110001010b

000000000010010x

AXBX

...

+ -

PC

Register

ALU

Take the data stored in memory address a, and move it to register AXTake the data stored in memory address b, and add it to register AXTake the data stored in register AX, and move it to memory address x

組合語言程式 /機器碼看到的電腦，

離實際電腦還是有一段差距e.g. 多核

心、 hyperthreading

當然仍需要再一或多層的轉換！

Computer Model of a Lower Layer

PC

Instructionmemory

Inst

ruct

ion

Add

Instruction[20– 16]

Mem

toR

eg

ALUOp

Branch

RegDst

ALUSrc

4

16 32Instruction[15– 0]

0

0

Mux

0

1

Add Addresult

RegistersWriteregister

Writedata

Readdata 1

Readdata 2

Readregister 1

Readregister 2

Signextend

Mux

1

ALUresult

Zero

Writedata

Readdata

Mux

1

ALUcontrol

Shiftleft 2

RegW

rite

MemRead

Control

ALU

Instruction[15– 11]

6

EX

M

WB

M

WB

WBIF/ID

PCSrc

ID/EX

EX/MEM

MEM/WB

Mux

0

1

Mem

Write

AddressData

memory

Address

(from Computer Architecture textbook)

11

A Layered View of Computer

+

if for

&&CPU

Memory

i, j, k;xfloat, yfloat;

A[0], A[1], …

i = i + j;if (A[0]==0)…

ProgramCounter

CPU

MemoryADD AX, bMOV x, AX… 010100110010101a

110010110001010b

000000000010010x

AXBXJX

...

+ -

PC

PC

Instructionmemory

Instruction

Add

Instruction[20– 16]

MemtoReg

ALUOp

Branch

RegDst

ALUSrc

4

16 32Instruction[15– 0]

0

0

Mux

0

1

Add Addresult

RegistersWriteregister

Writedata

Readdata 1

Readdata 2

Readregister 1

Readregister 2

Signextend

Mux

1

ALUresult

Zero

Writedata

Readdata

Mux

1

ALUcontrol

Shiftleft 2

RegW

rite

MemRead

Control

ALU

Instruction[15– 11]

6

EX

M

WB

M

WB

WBIF/ID

PCSrc

ID/EX

EX/MEM

MEM/WB

Mux

0

1

MemWrite

AddressData

memory

Address

i = i + j;xfloat = 1.0;if (A[0]==0)…

MOV AX, aADD AX, bMOV x, AX…

xxxxxxxxxxx

12

What’s Next?

Virtual machine concept (Sec. 1-2) Data representation (Sec. 1-3)

13

Virtual Machine Concept

Purpose of this section: Understand the role of assembly language in a

computer system

Side product: The principle of layered abstraction for combating

complexities, e.g. OSI 7-layer protocol

14

Virtual Machine Concept

A layered abstraction ofcomputers proposed byA. Tanenbaum

Each layer provides anabstract computer, orvirtual machine, to itsupper layer

Virtual machine: A hypothetical computer

that can be constructedof either HW or SW

What is a computer?

15

Simplest Model of Computers

Computeengine

Memory

Input data

Output data

ProgramInstructions

c.f., y = f(x)

Layered abstraction: A computer consists of layers of such virtual machine abstractions

16

Why Layered Abstraction?

Big idea: layered abstraction to combat complexities A strategy of divide-and-conquer Decompose a complex system into layers with

well-defined interfaces Each layer is easier to manage and handle Only need to focus on a particular layer, e.g. to

make it the best Also, it makes interaction clear

Particularly if one layer is realized in hardware and the other in software

17

Layered Abstraction of Computer

Each layer as a hypothetical computer, or virtual machine, that runs a programming language Can be programmed with the programming

language to process inputs and outputs

Program written in Li canbe mapped to that Li-1 by: Interpretation: Li-1 program interprets and executes

Li instructions one by one Translation: Li program is completely translated into

Li-1 program, and runs on Li-1 machine

Computeengine

MemoryInput data

Output data

ProgramInstructions

18

Layered Abstraction of Computer

+

if for

&&CPU

Memory

i, j, k;xfloat, yfloat;

A[0], A[1], …

i = i + j;if (A[0]==0)…

ProgramCounter

CPU

MemoryADD AX, bMOV x, AX… 010100110010101a

110010110001010b

000000000010010x

AXBXJX

...

+ -

PC

PC

Instructionmemory

Instruction

Add

Instruction[20– 16]

MemtoReg

ALUOp

Branch

RegDst

ALUSrc

4

16 32Instruction[15– 0]

0

0

Mux

0

1

Add Addresult

RegistersWriteregister

Writedata

Readdata 1

Readdata 2

Readregister 1

Readregister 2

Signextend

Mux

1

ALUresult

Zero

Writedata

Readdata

Mux

1

ALUcontrol

Shiftleft 2

RegW

rite

MemRead

Control

ALU

Instruction[15– 11]

6

EX

M

WB

M

WB

WBIF/ID

PCSrc

ID/EX

EX/MEM

MEM/WB

Mux

0

1

MemWrite

AddressData

memory

Address

i = i + j;xfloat = 1.0;if (A[0]==0)…

MOV AX, aADD AX, bMOV x, AX…

xxxxxxxxxxx

Li

Li-1

Virtual Machine

19

Languages of Different Layers

English: Display the sum of A times B plus C.

C++: cout << (A * B + C);

Assembly Language:

mov eax,Amul Badd eax,Ccall WriteInt

Intel Machine Language:

A1 00000000

F7 25 00000004

03 05 00000008

E8 00500000

20

High-Level Language

Level 5 Application-oriented languages, e.g., C, C++,

Java, Perl Written with certain programming model in mind

Variables in storage Operators for operations

Programs compiled into assembly language (Level 4) or interpreted by interpreters

What kind of computer does C see?

21

Assembly Language

Level 4 Instruction mnemonics that have a one-to-one

correspondence to machine language Based on a view of machine: register

organization, addressing, operand types and locations, functional units, …

Calls functions written at theOS level (Level 3)

Programs are translated intomachine language (Level 2)

What kind of computer does it see?

22

Operating System

Level 3 Provides services to Level 4 programs as if it

were a computer

How about VMware? How about JVM?

23

Instruction Set Architecture

Level 2 Known as conventional machine language Attributes of a computer as seen by assembly

programmer, i.e. conceptual structure and functional behavior Organization of programmable storage Data types and data structures Instruction set and formats Addressing modes and data accessing

Executed by Level 1 program (microarchitecture)

24

Microarchitecture

Level 1 Can be described by register transfer language

(RTL) Interprets conventional machine instructions

(Level 2) Executed by digital hardware (Level 0)

ZN

Register

Memory

PCIR

Controller ALU

clock

Control Signals

25

Digital Logic

Level 0 CPU, constructed from digital logic gates System bus Memory

26

What’s Next?

Virtual machine concept (Sec. 1-2) Data representation (Sec. 1-3)

27

Data Representation

Purpose of this section Assembly program often needs to process data,

and manage data storage and memory locations need to know data representation and storage

Binary numbers: translating between binary and decimal

Binary addition Integer storage sizes Hexadecimal integers: translating between

decimal and hex.; hex. subtraction Signed integers: binary subtraction Character storage

28

Integer Storage Sizes

What is the largest unsigned integer that may be stored in 20 bits?

Standard sizes:

Why unsignednumbers?

Why unsignednumbers?

29

Signed Integers

The highest bit indicates the sign1 = negative, 0 = positive

If the highest digit of a hexadecimal integer is > 7, the value is negative. Examples: 8A, C5, A2, 9D

30

Forming Two's Complement

Negative numbers are stored in two's complement notation Complement (reverse) each bit Add 1

Note that 00000001 + 11111111 = 00000000

Why?Why?

31

Binary Subtraction

When subtracting A – B, convert B to its two's complement

Add A to (–B)

1 1 0 0 1 1 0 0– 0 0 1 1 + 1 1 0 1

1 0 0 1

Practice: Subtract 0101 from 1001

32

Ranges of Signed Integers

The highest bit is reserved for the sign This limits the range:

What is the largest positive value that may be stored in 20 bits?

33

Character Storage

Character sets Standard ASCII (0 – 127)

‒ American Standard Code for Information Interchange

Extended ASCII (0 – 255) ANSI (0 – 255)

‒ American National Standard Institute Unicode (0 – 65,535): UTF-8. UTF-16, UTF-32

Null-terminated string Array of characters followed by a null byte

Using the ASCII table back inside cover of book

34

Chapter Summary

Assembly language helps you learn how software is constructed at the lowest levels

Assembly language has a one-to-one relationship with machine language

Each layer in a computer's architecture is an abstraction of a machine layers can be hardware or software

Assembly programs and assemblers often need to manage data storage in memory, and it is important to understand data representation and size of different data