川大软件学院左航 Now we begin our exploration in the assembly language.

川大软件学院左航

Now we begin our exploration

in the assembly language


Chapter 4

Chapter 4:Assembly Language-

Addressing Modes( 寻址方式)


Chapter 4:

• 4.1 Prerequisite knowledge about instructions

• 4.2 Data Addressing Modes

• 4.3 Program Memory-Addressing Modes

• 4.4 Stack Memory-Addressing Modes


4.1 Prerequisite knowledge about

instructions


4.1 Prerequisite knowledge about instructions

• Default expression– Segment register ---- stored segment base

address– imm/im ---- immediate data– DST ---- destination operand– SRC ---- source operand– R8/R16/R32----register of 8/16/32 bits– mm ---- memory data



• MOV AX, BX

opcode operand operand

1,2,3 operand SS/DSCS



• We will use the most common instruction “MOV” to illustrate data addressing modes.

MOV operand1 operand2

Copy content

DST SRC



• MOV X,Y X = Y

opcode operand1 operand2

AX

BX

CX

DX

00100101

11001011

11110000

10111001

10101110

immediate data MOV AX, 23H

2.register

3.memory

1.immediate data



• 1. Operand is directly in the instructions.

• ----immediately addressing

• 2. Operand is stored in a register in CPU.

• ----register addressing

• 3. Operand is stored in the memory.• ----the address is combined by segment

address and offset address.


4.2 Data Addressing Modes


4.2 Data Addressing Modes• 4.2.2 Immediate addressing• 4.2.3 Direct data addressing• 4.2.1 Register addressing• 4.2.4 Register indirect addressing• 4.2.6 Register relative addressing• 4.2.5 Base-plus-index addressing• 4.2.7 Base relative-plus-index

addressing• 4.2.8 Scaled-index addressing• 4.2.9 Data Structures


4.2.2 Immediate addressing



• MOV AX,1234H AX = AX + 1234H


AX

BX

CX

DX

00100101

11001011

11110000

10111001

10101110

immediate data 1234H

memcpu


12H34HOpecode

00000H

FFFFFH

1FFFFH

3FFFFH

90000H

9FFFFH

MEMMORY 1M

CS

30000HDS

SS 10000H

CS + IP

zuohang

Review:Because some time the students can't understand the addressimg mode clearly.



• Example:– MOV AX,1234H

AH AL

12H

34H

opcode

AX

Code segment

00105

00104

00103



• Definition:– Operand2 is a immediate data or an

expression whose value can be figure out. It’s a constant data.

• Example:– MOV AX,100H AX ← 100H

– MOV EAX,0A5FH EAX ←0A5FH

– MOV AH,1101B AH ←1101B

– MOV AL,’A’---------AL ← 41H


4.2.2 Immediate addressing• A simple program .MODEL TINY (only cs)

0000 .CODE .STARTUP0100 B8 0000 MOV AX,00103 B8 0000 MOV BX,0000H0106 B9 0000 MOV CX,00109 8B F0 MOV SI,AX .EXIT (return to DOS) END


4.2.3 Direct data addressing



• MOV AX,NUM AX = AX + NUM


AX

BX

CX

DX

00100101

11001011

11110000

10111001

10101110

immediate data 1234H


20H00HOpecode

3050

00000H

FFFFFH

1FFFFH

3FFFFH

90000H

9FFFFH

MEMMORY 1M

CS

30000HDS

SS 10000H

Physical address

= segment base address + offset

=DS + 2000H

Address of NUM

Content of NUM

zuohang

Review:Because some time the students can't understand the addressimg mode clearly.



• A. Direct addressing• B. Displacement Addressing



• A. Direct addressing– Definition: Transfer data between a

memory location, located within the data segment and the AX register. P79

– Example:• MOV AX, NUM AX ←DS:[NUM]

• MOV TWO,AX DS:[TWO] ←AX

• MOV ES:[2000H],AL ES:[2000] ←AL



30 00 020

00

opcode

30

50

AH ALAX

DS

30002

Code

segment

data

segment

30000

2000

32000

+

32001

32000

2

3

1

4

NUM DW 3050H

MOV AX,NUM

NUM

zuohang

All the segments are infact in one memory, but we divide the memory into several segments.The address is the physical address.



• SS+SP/ESP or SS+BP/EBP

• DS+ memory offset

• MOV ES:[2000H],AL ES:[2000] ←AL



• B. Displacement Addressing– It is almost identical to direct

addressing, except that the instruction is four bytes wide instead of three.

– And the registers used aren’t AX.• MOV AX, NUM AX ←DS:[NUM]

– We need not consider about the instruction bytes now, so ignore it.• MOV CX,NUM


4.2.1 Register addressing


4.2.1 Register addressing

• Definition:– All the operands are in

registers(8/16/32b). It’s a variable data.

• Example:– MOV AL,BL (8) AL ← BL

– MOV DS,AX (16) DS ← AX

– MOV SP,BP (16) SP ← BP

– MOV ECX,EDX ECX ←EDX


4.2.1 Register addressing• Special Example:

– MOV EAX,BX– Wrong, mixed size in MOV (32 ,16)

– MOV DS,AX– Right

– MOV ES,DS– Wrong, segment-to-segment

– MOV CS,BX or MOV IP,BX– Wrong, CS or IP Register can’t be DST


4.2.4 Register indirect addressing

• MOV AX,BX• MOV AX,[BX]


4.2.4 Register indirect addressing• Definition:

– physical address = (BP,BX)/(DI,SI) + base address (segment registers)

– MOV AX,[BX]

– memory data– offset address is stored BX 、 SI 、 DI 、

BP.– [ ] means indirect addressing– MOV AX,[DX] wrong



2 0 0 0 0

X X

opcode

30

50AH ALAX

DS

20002

Code

segment

data

segment

20000

1000

21000

+

21001

21000

1

2

3

MOV AX,[SI]

1 0 0 0 SI

zuohang

if one operand is the data in memory ,pay special attention to the other operand , which indicated the size of the data in memory



• A. DS --------[BX] 、 [SI] 、 [DI]

• Example:

MOV AX ， [BX] AX← (DS:[BX])

MOV AH ， [DI] AH← (DS:[DI])

Base address + offset

zuohang

usually we use DST operand to judge the addressing model



• B. SS-------- [BP]

• Example:

MOV AX ， [BP] AX← (SS:[BP])

• MOV ES:[2000H],AL [ES:[2000]] ←AL

Base address + offset



• Example: MOV [DI],10H cause ambiguous

DI = 0200H,DS =1000HMOV BYTE PTR [DI],10H

MOV WORD PTR [DI],10H

00H

10H

10H

10200H

10200H

10201H



• MOV [DI],[BX]– Wrong , memory to memory is not

permitted.


4.2.4 Register indirect addressing• Simple program P83 3-6

.MODEL SMALL (DATA & CODE)

.DATA AGAIN: DATAS DW 50 DUP (?) MOV AX,ES:

[046CH] .CODE MOV [BX],AX .STARTUP INC BX

INC BX MOV AX,0 LOOP AGAIN MOV ES,AX .EXIT MOV BX,OFFSET DATAS END MOV CX,50


?

?

? 10000H

10001H

10031H

10032H

DB 50 DUP (?)

Data Segment

DS = 1000H

DATAS = 0


4.2.6 Register relative addressing



•Definition:– physical address = displacement

( 位移量 ) + (BP,BX)/(DI,SI) + base address (segment registers)

•Examples:– MOV AX,[SI+100H] AX ←DS:

[SI+100H]



• MOV AX,[SI+100H]

30 00 0

X X

opcode

30

50

Code

segment

data

segment

1

00

05 00

30600

AH ALAX

DS

SI

30601



•Examples:– MOV AX,[DI+100H] AX ←DS:

[DI+100H]– MOV ARRAY[SI],BL DS:[ARRAY+SI] ←BL– MOV LIST[SI+2],CL DS:[LIST+SI+2] ←CL

– MOV CX,[BP+10H] CX ←SS:[BP+10H]



• Simple program p88 3-8

• .MODEL SMALL MOV DI,10H

• .DATA MOV AL,ARRAY[DI]

• ARRAY DB 16 DUP (?) MOV DI,20H

• DB 29H MOV ARRAY[DI],AL

• DB 30 DUP (?) .EXIT

• .CODE END

• .STARTUP


? 17 29H 16 ? 15

? 0

FFFFFH

00000H

DS = 30000H

ARRAY DB 16 DUP (?)

OFFSET

DB 30 DUP (?)

OFFSET = 0FH = ARRAY + 15

ARRAY OFFSET = 0

OFFSET = 10H = ARRAY + 16

DS + 01H

DB 29H

P85 3-7

zuohang

Specialy it explain the example on page 88 3-8Like in the c language the array point to the first element.


29H 00020H

?

?

29H

?

?

00000H

00001H

0000FH

00010H

00011H

0002EH

DB 16 DUP (?)

DB 30 DUP (?)

DB 29HData Segment

DS =0000H

ARRAY = 0

MOV AL, ARRAY[DI] ----ARRAY+DI+DS*10H


4.2.5 Base-plus-index addressing



• Definition:– memory data.– offset = BP/BX( 基址 )+ DI/SI ( 变址 )– Physical address = offset + DS/SS

• Examples: MOV CX,[BX+DI] CX ←DS:[BX+DI] 16

MOV CH,[BP+SI] CH ←SS:[BP+SI] 8



• MOV AX ，[BX+SI]

30 00 0

X X

opcode

30

50

Code

segment

data

segment

12 00

05 00

31700

AH ALAX

DS

BX

SI

31701



•Examples:MOV [BX+SI],SP DS:[BX+SI] ← SP

16MOV [BP+DI],AH SS:[BP+DI] ←AH

8MOV CL,[EDX+EDI] CL ←DS:[EDX+EDI]

8MOV [EDX+EDI],ECX DS:[EDX+EDI]←ECX

32

zuohang

some instructions are used in 80386 and above whose address bus is 32-bit wide. Others are used in 8086-80286,whose address bus is 8/16-bit wide.


4.2.5 Base-plus-index addressing• Simple program p85 3-7

• .DOMEL SMALL MOV BX, OFFSET ARRAY

• .DATA MOV DI,10H• ARRAY DB 16 DUP (?) MOV AL,[BX+DI]• DB 29H MOV DI,20H• DB 30 DUP (?) MOV [BX+DI],AL• .CODE .EXIT• .STARTUP END

• DS+OFFSET ARRAY + 10H


?

?

?

29H

?

? 00000H

00001H

0000FH

00010H

00011H

0002EH

DB 16 DUP (?)

DB 30 DUP (?)

DB 29HData Segment

DS =0000H

ARRAY = 0

DS+OFFSET ARRAY + 10H


4.2.7 Base relative-plus-index addressing



•Definition:– offset = displacement ( 位移量 ) +BP/BX

+DI/SI

– Physical address = offset + DS/SS

– Examples:– MOV DH,[BX+DI+100H] DH DS:

[BX+DI+100H]



• MOV AX,[BX+DI+100H]

30 00 0

X X

opcode

30

50

Code

segment

data

segment

1

00

05 00

31700

AH ALAX

DS

DI11 00BX

31701



• Examples:– MOV DH,[BX+DI+20H] DH ←DS:

[BX+DI+20H]

– MOV AX,FILE[BX+DI] AX ←DS:[FILE+BX+DI]

– MOV LIST[BP+DI],CL SS:[LIST+BP+DI] ←CL– MOV EAX,FILE[EBX+ECX+2]

• Simple program p89 3-9


• .MODEL SMALL• .DATA• FILE EQU THIS BYTE MOV BX, OFFSET

RECA• RECA DB 10 DUP (?) MOV DI, 0• RECB DB 10 DUP (?) MOV AL, FILE[BX+DI]• RECC DB 10 DUP (?) MOV BX, OFFSET

RECC• RECD DB 10 DUP (?) MOV DI, 2• .CODE MOV FILE[BX+DI], AL• .STARTUP .EXIT • END

Variable defined by EQU does not take up space in memory, but it does represent type information.

zuohang

equ 定义的变量不占用存储空间，但使用其作为偏移量时，注意其类型必须和另外一个操作数匹配。并且可以当作变量名使用，比如MOV AL,FILE是正确指令。MOV AL,FILE+1,也是正确指令。


• .1EH

1D Ｈ

15Ｈ14Ｈ

０ B Ｈ０ A Ｈ

０ 0 Ｈ０ 0 ＨOFFSET RECA = 0 Low

address

High address

RECA

10

10

11

2021

3031

OFFSET RECB = 10

OFFSET RECC =20

OFFSET RECC =30

RECB

RECC

RECD


4.2.8 Scaled-index addressing



• Definition:– It’s used in 80386 and above. Designed

to address word and doubleword more easily.

– Physical address = (32-bit base register) + scaling factor * (index registers)

• Example:– MOV EAX,[EBX+4*ECX] ---- doubleword– MOV AX,[EBX+2*ECX] ---- word



• P91 　　３－１０

• .MODEL SMALL MOV EBX, OFFSET LIST• .386 MOV ECX, 2• .DATA MOV [EBX+2*ECX], AX• LIST DW 0,1,2,3,4 MOV ECX, 4• DW 5,6,7,8 MOV [EBX+2*ECX], AX• .CODE INC ECX• .STARTUP INC ECX• MOV [EBX+2*ECX], AX• .EXIT• END



• Simple program P91 　　３－１００ 0 Ｈ

０ 4 Ｈ０ 0 Ｈ０ 3 Ｈ０ 0 Ｈ０ 2 Ｈ０ 0 Ｈ０ 1 Ｈ０ 0 Ｈ０ 0 Ｈ

03

……

RIGHT

Low address

High address

OFFSET

04

020100

05

06

07

08


Summary of data addressing


Summary of data addressing 1.Register addressing

– MOV AX,BX

34H 12H 34H 12H

AX BX


4.Register indirect addressing

2 0 0 0 0

X X

opcode

30

50AH ALAX

DS

20002

Code

segment

data

segment

20000

1000

21000

+

21001

21000

1

2

3

MOV AX,[BX]

1 0 0 0 BX

zuohang

if one operand is the data in memory ,pay special attention to the other operand , which indicated the size of the data in memory


6.Register Relative Addressing

• MOV AX,[BX+100H]

30 00 0 X X

opcode

30

50

Code

segment

data

segment

1

00

05 00

30600

AH ALAX

DS

BX


Summary of data addressing 2.Immediate addressing

– MOV CX,1234H

12H 34H12H

CX 34H

OPCODE

CS


Summary of data addressing3.Direct data addressing

– MOV AX,NUM

30H 50H30H

AX50H

DS 30000H

2000H ?

DS

NUM

NUM32000H

32001H


30 00 0 20

00

opcode

30H

50HAH ALAX

DS

30002

Code

segment

data

segment

30000

2000

32000

+

32001

32000

2

3

1

4

NUM DW 3050H

MOV AX,NUM

NUM

NUM = 2000H

zuohang

All the segments are infact in one memory, but we divide the memory into several segments.The address is the physical address.


5.Base-plus-Index Addressing

• MOV AX ，[BX+SI]

30 00 0 XX

opcode

30

50

Code

segment

data

segment

12 00

05 00

31700

AH ALAX

DS

BX

SI

31701


7. Base relative-plus-index addressing

• MOV AX,[BX+DI+100H]

30 00 0

X X

opcode

30

50

Code

segment

data

segment

1

00

05 00

31700

AH ALAX

DS

DI11 00BX

31701


Summary of data addressing

• DS-----------BX,DI,SI• SS------------BP

• MOV AX,ARRAY[BX]• MOV AX,[BX+100H]


4.2.9 Data Structures



• If we want store an address book. Must we copy 100 times?

• Can we do it more efficiently?



• Solution:– If all the data is in the same format.

We can use data structure to define a template. It will do the duplicate work for us .



• Structure in c++– struct Info– {– char names[32];– char street[32];– char city[16];– };– void main()– {– Info classmates;– classmates = {“Frank”, “Wenhua road”,

“CD”};– }



• Example3.11 p91 INFO STRUC

NAMES DB 32 DUP(?)STREET DB 32 DUP(?)

CITY DB 16 DUP(?) INFO ENDS

NAME1 INFO <‘BOB’,’123STREET’,’CHENGDU’>

MOV SI,OFFSET NAME1.NAMESP93—3-12



• Example3.12 • INFO STRUC NAMES DB 32 DUP(?)

STREET DB 32 DUP(?) CITY DB 16 DUP(?) INFO ENDS

MOV CX,32MOV AL,0MOV DI,OFFSET NAME1.NAMESREP STOSB


4.3 Program Memory-Addressing Modes


4.3 Program Memory-Addressing Modes

• 4.3.1 Direct Program Memory Addressing

• 4.3.2 Relative Program Memory Addressing

• 4.3.3 Indirect Program Memory Addressing


4.3.1 Direct Program Memory Addressing


4.3.1 Direct Program Memory Addressing

• Definition:– the address is stored with the opcode.

• Example:– JMP [10000H]– CS:1000H;IP:0000H


4.3.2 Relative Program Memory Addressing

• Example:– JMP [2] p94 figure 3-15– Only IP is changed.

– 10000 EB– 10001 02– 10002 --– 10003 --– 10004

JMP [2]

IP before JMP

IP after JMP


4.3.3 Indirect Program Memory Addressing

• Example:– See p94 first para & p95 3-16/3-13– Only IP is changed.

– JMP AX– JMP NEAR PTR [BX]– JMP NEAR PRT[DI+2]– JMP TABLE[BX]– JMP ECX


4.3.3 Indirect Program Memory Addressing

– See p95 3-16/3-13

• Table DW LOC0• DW LOC1• DW LOC2

• MOV BX, 4• JMP TABLE[BX]


4.4 Stack Memory-Addressing Modes



• It is usually used in PUSH, POP, CALL to protect some important data or pointers.



• Example:– PUSH BX (LIFO)

12 34

EAX

EBX

ECX

EDX

ESP

SS*10H

XX

opcode

12 SP-1

34 SP-2

STACK

SEGMENT

CODE

SEGMENT

P96 2 para from bottom

Stack top

Stack bottom

FFFFFH

00000H


• We will discuss JMP, CALL, PUSH and POP in detail in the following chapters.

川大软件学院左航 Now we begin our exploration in the assembly language.

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Transcript of 川大软件学院左航 Now we begin our exploration in the assembly language.