addressing modes&instructions

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8086 Addressing Modes



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What is the Addressing Mode

?add dest, source ; dest +source→dest

add ax,bx ; ax +bx→ax

The addressing mode means where and how

the CPU gets the operands when theinstruction is executed.



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• Addressing modes for Sequential Control

Transfer Instructions----These Instructions transfer control to the

next sequential instruction in the program

• Addressing modes for Control TransferInstructions

-----These Instructions transfer control to

some predefined address Ex:INT CALL



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Addressing modes for Sequential Control

Transfer Instructions Three types of 8086 addressing modes

• Immediate Addressing Mode

---CPU gets the operand from the instruction • Register Addressing Mode

---CPU gets the operand from one of the internal registers

• Memory Addressing Mode---CPU gets the operand from the memory location(s)



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Exp

MOV AL, 80H

Machine code:B080H

ALB0H

80H

Instruction Queue

MACHINE

CODE

B8

12H

Instruction Queue

AL

MACHINE

CODE

AH

34H

12

3480H

80H

12 34

1. Immediate Addressing Mode

MOV AX, 1234H

Machine Code:B83412H



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Exp：MOV AX, CX

89

C1

Memory

AX

CX Machine code

2. Register Addressing

Mode



• Specify an offset address (effective address) using expressions of the form (different parts of expression are optional):

– [ Base Register + Index Register+ Displacement]

• 1) Base Register---BX, BP• 2) Index Register---SI, DI

• 3) Displacement ---constant value

• Example: 1) add ax,[20h] 2) add ax,[bx]

3) add ax,[bx+20h] 4) add ax, [bx+si]5) add ax, [bx+si+20h]

3. Memory Addressing Mode



⑴ Direct Addressing Mode

Exp: MOV AL, [1064H]

Machine code:A06410H

• The offset address of the operand is provided in theinstruction directly;

• The physical address can be calculated using thecontent of DS and the offset :

PA = (DS)*10H+Offset

3. Memory Addressing Mode



⑴ Direct Addressing ModeExample: MOV AL, [1064h] ;Assume (DS)=2000H

Machine code: A06410H

21064H

（DS)*10H=20000H

20000H

21064H

AL

A0

64

10

45

…

Code

Segment

Data

Segment

45

45

+ 1064H



⑵ Register Indirect Addressing Mode

• The address of memory location is in a register

(SI,DI,or BX only)

• The physical address is calculated using the content of DS

and the register(SI,DI,BX)

PA = (DS)*10H+(SI)/(DI)/(BX)

3. Memory AddressingMode



50

40

… …

M

AX

⑵ Register Indirect Addressing Mode

ASSUME: (DS)=3000H, (SI)=2000H, (BX)=1000H

30000H

(DS)*10H=30000H

(SI)= 2000H+

32000H

32000H

40 50

50

40

… … 64H

M

AL 30000H

(DS)*10h= 30000H

(BX)= 1000H+

31000H

31000H64H

64H

MOV [BX], ALMOV AX, [SI]



⑶ Register Relative Addressing

EA=

(BX)

(BP)

(DI)

(SI)

+ Displacement

For physical address calculation:DS is used for BX,DI,SI;SS is used for BP

PA=(DS)*10H+(BX)/(DI)/(SI)+Disp

ORPA=(SS)*10H+(BP)+Disp



⑶ Register Relative AddressingMOV CL, [BX+1064H] ;assume: (DS)=2000h, (bx)=1000h

;Machine Code: 8A8F6410

22064H

22064H

8F

64

10

45

…

CodeSegment

Data

Segment

8A

…

CL

45

45 21000H

(BX)= 1000H

(DS)*10h= 20000H

20000H

+ 1064H

PA=(ds)*10h+(bx)+1064h



⑷ Based Indexed Addressing

EA=(BX)(BP)

+(DI)(SI)

• Base register(bx or bp) determines which segment(data or

stack) the operand is stored;• if using BX, the operand is defaultly located in Datasegment,then:

PA=(DS)*10H+(BX)+(DI)/(SI)

• if using BP, the operand is defaultly located in stack segment,then:

PA=(SS)*10H+(BP)+(DI)/(SI)



⑷ Based Indexed Addressing

Example: MOV AH, [BP][SI];Assume(ss)=4000h,(bp)=2000h,(si)=1200h

56H

…

…

M

AH 40000H

(SS)*10H= 40000H

(BP)= 2000H+

43200H

43200H

(SI)= 1200H

56H

56H

PA=(ss)*10h+(bp)+(si)



⑸ Based Indexed Relative Addressing

EA= (BX)(BP)

+ (DI)(SI)

+ Displacement

if using BX, the operand is defaultly located in Datasegment,then:

PA=(DS)*10H+(BX)+(DI)/(SI)+disp

if using BP, the operand is defaultly located in stack segment,then:

PA=(SS)*10H+(BP)+(DI)/(SI)+disp



⑸ Based Indexed Relative Addressing

MOV [BX+DI+1234H], AH;assume (ds)=4000h,(bx)=0200h,(di)=0010h

;machine code:88A13412hA1

34

12

…

Code

segment

Data

segment

88

…

45AH

40000H

(DS)*10H=40000H

(BX)= 0200H

+(DI)= 0010H

1234H

45

45

41444H

41444H



Summary on the 8086 memory addressing modes

operand offset address Default Overridden

（effective address） Segment Register Segment Register

3. Register [SI/DI/BX/BP+disp] (SI)/(DI)/(BX)/(BP)+disp DS CS ES SS

Relative Addressing

2. Register [BX]/[SI] /[DI] Content of the R DS CS ES SS

Indirect Addressing

1. Direct Addressing [disp] disp DS CS ES SS

4. Based Indexed [BX+SI/DI] (BX)+disp DS CS ES SS

Addressing [BP+SI/DI] (BP)+disp SS CS ES DS

5. Based Indexed [BX+SI/DI+disp] (BX)+(SI)/(DI)+disp DS CS ES SS

Relative Addressing [BP+SI/DI+disp] (BP)+(SI)/(DI)+disp SS CS ES DS



Examples:

Assume: (BX)=6000H, (BP)=4000H, (SI)=2000H,(DS)=3000H, (ES)=3500H, (SS)=5000H

3000:0520 30520HDirect Addressing

2. MOV AX, [BX]

1. MOV AX, [0520H]

5. MOV AX, ES: [BX+SI+0050H]

4. MOV AX, [BP+6060H]

Register Indirect Addressing 3000:6000 36000H

Register Relative Addressing

Register Relative Addressing

3. MOV AX, [SI+1000H] 3000:3000 33000H

5000:A060 5A060H

3500:8050 3D050HBased Indexed Relative

Addressing

Instruction addressing logical physicalmode address address



•Addressing modes for Control Transfer

Instructions

Modes for ControlTransfer Instructions

Inter-segment

Intra-segment

Inter-segment-In Direct

Inter-segment-Direct

Intra-segment Direct

Intra-segment Indirect



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Inter-segment-Direct Addressing ModeThe address to which control is to be transferred lies in the different

segment and appears directly In the instruction as an immediatedisplacement value w.r.t IP If the displacement is8-bits(-128<d<+127)(short) If the displacement is16-bits(-32768<d<+32767)(Long)

Ex: CALL 0020:0010H

segment1

CS

Assume CS=2000h,IP=0000h

After JMP,CS= 0020h,IP=0010h

segment2

IP

20

00

10

Sub-routine

…

00

…

0020

Op-code for CALL

0010

2000hCS



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Inter-segment-IN-Direct Addressing ModeThe address to which control is to betransferred

lies in the different \segment and it ispassed to theinstruction indirectly i.e contents of amemory blockcontaining four bytesIP(LSB),IP(MSB),CS(LSB),CS(MSB)

Ex: CALL [BX]

IP(LSB)10

IP(MSB)00

CS(LSB)20

Sub-routine

…

segment1

…

0020CS

Before JMP,Assume BX=100h, CS=3000h,IP=200h

Op-code for CALL

After JMP,CS= 0020h,IP=0010h

segment2

0010IP

3000hCS

CS(MSB)00

0010IP



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Intra-segment-Direct Addressing ModeThe address to which control is to be transferred lies in the samesegment and appears directly In the instruction as an immediate

displacement value w.r.t IP If the displacement is8-bits(-128<d<+127)(short) If the displacement is16-bits(-32768<d<+32767)(Long)

Ex: CALL 500h

CodeSegment

Assume CS=2000h,IP=0000h

After CALL,CS= 2000h,IP=IP+500h

IP

00

05

Op-code for CALL2000hCS

0500Sub-routine

0000IP



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IP

Intra-segment-In-Direct Addressing ModeIn this mode the displacement to which control is to be transferred,Is in the same segment in which the control transfer instruction lies

But it is passed to the instruction indirectly

Ex: CALL [BX]

CodeSegment

Assume CS=2000h,IP=0000h , BX=800h

After CALL,CS= 2000h,IP=IP+800h

IP

00

05

Op-code for CALL2000hCS

0800Sub-routine

0000IP



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Example:The Contents of different registers are given below. Form effective addresses fordifferent addressing modes

Offset(displacement)= 5000HAX=1000H,BX=2000H,SI=3000H,DI=4000H,BP=5000H,SP=6000H,CS=0000H,DS=1000HSS=2000H,IP=7000HShifting a number four times is equivalent to multiplying it by 16D or 10H



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Instruction Set

&Assembler Directives



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Programming in 8088/8086 Three levels of languages available to program a microprocessor:

Machine Languages , Assembly Languages , and High-level

Languages .

Machine Language

A sequence of binary codes for the instruction to be executed by

microcomputers.

Long binary bits can be simplified by using Hexadecimal format

It is difficult to program and error prone.

Different uP (micro-processor) uses different machine codes.



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Programming in 8088/8086 (cont.)

Assembly Language

To simplify the programming, assembly language (instead of machine language) isused.

Assembly language uses 2- to 4-letter mnemonics to represent each instructiontype. E.g. “Subtraction” is represented by SUB

Four fields in assembly language statement:

Label , OP Code , Operand and Comment fields.

Programs will be „translated‟ into machine language, by Assembler, so it can beloaded into memory for execution.

High-Level Language

High level languages, like C , Basic or Pascal , can also be used to program

microcomputers.

An interpreter or a compiler is used to „translate‟ high level language statement tomachine code. High level language is easier to read by human and is more suitablewhen the programs involves complex data structures.

Assemblers



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Assemblers

Programming the instructions directly in machine code is possible butevery machine codes depending on how the data is stored.

The process of converting the microprocessor instructions to thebinary machine code can be performed automatically by a computerprogram, called an ASSEMBLER. Popular assemblers include IBMmacro Assembler, Microsoft Macro Assembler (MASM) and BorlandTurbo Assembler(installed on IE NT Network).

Most assemblers accept an input text file containing lines with arigorously defined syntax split into four fields.

Not all fields need to be present in a line. Eg. A line can be just acomment line if it starts with semicolon;

Source Codes Object Codes and Linking



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Source Codes, Object Codes and Linking

Source code is the text written by the programmerin assembly language

(or any other programming language)

Object code is the binary code obtained afterrunning the assembler (

Or compiler if the source is in a high level language).

Modules of a program may be written separatelyand linked together to

form a executable program using a linker.

The linker joins the object code of the different

modules into one largeobject file which is executable. Most assemblers onIBM PCs produce

object files which must be linked ( even if there areno separate modules).

Source Codes, Object Codes and Linking(Contd.,)



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Source Codes, Object Codes and Linking(Contd.,)

Fields in Assembler



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Fields in Assembler

<label> <Mnemonic or directive> <operands> <;comment>

Comment field contains internal program documentation to improve

human readability -use meaningful comments

Operand field contains data or address used by the instruction.

The following conventions typically apply:

Fields in Assembler (Contd )



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Fields in Assembler (Contd.,)

<label> <Mnemonic or directive> <operands><;comment>

Mnemonic/directive field contains the abbreviation forthe processor instruction (eg. MOV) or an assemblerDIRECTIVE. Adirective produces no object code but isused to control how the assembler operates.

Examples of directives include:

END -indicate the end of a program listing,

FRED LABEL NEAR - define “FRED” as a near label

TOM EQU 1000H -define “TOM” as the number 1000H

Label field contains a label which is assigned a valueequal to the address where the label appears.

Why Program in Assembler?



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Why Program in Assembler?

Assembler language instruction has a one-to-onecorrespondence with the binary machine code: theprogrammer controls precisely all the operations performedby the processor (a high level language relies on a compileror interpreter to generate the instructions).

Assembler can generate faster and more compactprograms

Assembler language allows direct access and full controlof input/output operations

However, high-level language programs are easier towrite and develop than assembler language programs

Advantages of High-level languages



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Advantages of High level languages

Block structure code: programs are most readable when

they arebroken into “logical blocks” that perform specific function.

Productivity: easier to program

Level of complexity: no need to know the hardware details

Simple mathematics formula statement

Portability: only need to change the compiler when it isported to other machine

Abstract data types: different data types like floating-pointvalue,

record and array, and high precision value.

Readabilit

Intel 8086 Instruction Set Overview



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Intel 8088 has ninety basic ( ie not counting

addressing mode

variants) instructions

Instructions belong to one of the following groups:data

transfer, arithmetic, logic, string manipulation,control

transfer and processor control.

C i A bl L I i



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Converting Assembly Language Instructions to

Machine Code

• An instruction can be coded with 1 to 6 bytes

• Byte 1 contains three kinds of information – Opcode field (6 bits) specifies the operation (add, subtract, move)

– Register Direction Bit (D bit) Tells the register operand in REG field in byte

2 is source or destination operand1: destination 0: source

-Data Size Bit (W bit) Specifies whether the operation will be performed on 8-

bit or 16-bit data

0: 8 bits 1: 16 bits

• Byte 2 has three fields



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• Byte 2 has three fields

– Mode field (MOD)

– Register field (REG) used to identify the register for the first operand

– Register/memory field (R/M field)



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2-bit MOD field and 3-bit R/M field together

specify the second operand

Mode Field encoding

Register/memory (R/M) Field Encoding



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Examples

MOV BL,AL (88C316)

Opcode for MOV = 100010

D = 0 (AL source operand)

W bit = 0 (8-bits)

Therefore byte 1 is 100010002=8816

• MOD = 11 (register mode)

• REG = 000 (code for AL)

• R/M = 011 (destination is BL)

Therefore Byte 2 is 110000112=C316



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Examples:

MOV BL, AL = 10001000 11000011 = 88 C3h

ADD AX, [SI] = 00000011 00000100 = 03 04 h

ADD [BX] [DI] + 1234h, AX = 00000001 10000001 __ __ h =

01 81 34 12 h




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I. Data Movement Instructions (14)



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(abbreviations below: d=destination, s=source)

General Data Movement Instructions

MOV d,s - moves byte or word; most commonly used instructionPUSH s - stores a word (register or memory) onto the stack

POP d - removes a word from the stack

XCHG d,s - exchanges data, reg.-reg. Or memory to register

XLAT - translates a byte using a lookup table (has no operands)

IN d,s - moves data (byte or word) from I/O port to AX or AL

OUT d,s - moves data (byte or word) from AX or AL to I/O port

LEA d,s - loads effective address (not data at address) into register

LDS d,s - loads 4 bytes (starting at s) to pointer (d) and DS

LES d,s - loads 4 bytes (starting at s) to pointer (d) and ES

LAHF - loads the low-order bytes of the FLAGS register to AH

SAHF - stores AH into the low-order byte of FLAGS

PUSHF - copies the FLAGS register to the stack

POPF - copies a word from the stack to the FLAGS register

( )

Instructions for moving strings



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g g

String instructions are repeated when prefixed by the REP mnemonic (CXcontains the repetition count)

MOVS d,s - (MOVSB, MOVSW) memory to memory data transfer

LODS s - (LODSB and LODSW) copies data into AX or AH

STOS d - (STOSB, STOSW) stores data from AH or AX

Data movement using MOV



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g

MOV d, s

d=destination (register or effective memory address),

s=source (immediate data, register or memory address) MOV can transfer data from:

any register to any register (except CS register)

memory to any register (except CS)

immediate operand to any register (except CS)

any register to a memory location

immediate operand to memory

MOV cannot perform memory to memory transfers (must use a register as anintermediate storage).

MOV moves a word or byte depending on the operand bit-lengths; the source

and destination operands must have the same bit length.

MOV cannot be used to transfer data directly into the CS register.

The stack The stack



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The stack is a block of memory reserved for temporary

storage of data and registers. Access is LAST-IN, FIRSTOUT (LIFO)

The last memory location used in the stack is given by theeffective

address calculated from the SP register and the SS register:

Example:

The stack



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Data may be stored onto the stack using the PUSH

instruction –this automatically decrements SP by 2 (allstack operations involve words).

The POP instruction removes data from the stack (andincrements SP by 2).

The stack may be up to 64K-bytes in length.

PUSH and POP instructions



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Examples:

PUSH AX ;stores AX onto the stack

POP AX ;removes a word from the stack and loads it into AX

PUSHF ;stores the FLAGS register onto the stack

POPF ; removes a word from the stack and loads it into FLAGS

PUSH may be used with any register to save a word (the register

contents) onto the stack. The usual order (e.g. as with MOV) of storing the lower order byte in the lower memory location is used.

PUSH may also be used with immediate data, or data in memory.

POP is the inverse of the PUSH instruction; it removes a word

from the top of the stack. Any memory location or 16-bit register

(except CS) may be used as the destination of a POP instruction.

PUSHF and POPF saves and loads the FLAGS register to/from the

stack,respectively.

Exchange Instruction (XCHG)



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XCHG exchanges the contents of two registers or a registerand memory. Both byte and word sized exchanges are

possible.

Examples:

XCHG AX,BX; exchange the contents of AX and BX

XCHG CL,BL; exchange CL and BL contents

XCHG DX,FRED; exchanges content of DX and memory

DS:FRED

Memory to Memory exchanges using XCHG are NOT allowed.

Translate Instruction (XLAT)



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Many applications need to quickly convert byte sized codes to other valuesmapping one byte value to another (e.g. mapping keyboard binary codes to ASCIIcode)

XLAT can perform a byte translation using a look-up table containing up to 256elements

XLAT assumes that the 256-byte table starts at the address given by DS:BX (i.e.effective address formed by the DS and BX registers). AL is used as an index topoint to the required element in the table prior to the execution of XLAT. The resultof XLAT instruction is returned in the same register (AL).

Address Data Table

LEA &LDS



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LEA loads the offset of a memory address into a 16-bit register. The offset addressmay be specified by any of the addressing modes.

Examples (with BP=1000H):

LEA AX,[BP+40H];=>SS:[1000H+40H] =SS:[1040H];load 1040H into AX

LEA BX,FRED; load the offset of FRED (in data segment) to BX

LEA CX,ES:FRED; loads the offset of FRED (in extra segment) to CX

LDS -Load data and DSLDS reads two words from the consecutive memory locations and loads them into thespecified register and the DS segment registers.

Examples (DS=1000H initially)

LDS BX,[2222H]; copies content of 12222H to BL, 12223H to BH, and 12224 and12225 to DS register

LDS is useful for initializing the SI and DS registers before a string operation. E.g.LDS SI, sting_pointer

The source for LDS can be displacement, index or pointer register (except SP).

LES -Load data and ES



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LES reads two words from memory and is very similar to LDS

except that the second word is stored in ES instead of DS.

LES is useful for initializing that DI and ES registers for stringsoperation.

Example (with DS=1000H):

LES DI, [2222H]; loads DI with contents stored at 12222H and12223H and loads ES with contents at 12224 and 12225H

LAHF, SAHF



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LAHF (load AH with the low-order byte of the FLAGSregister) and SAHF (Store AH into the low-order byte of theFLAG register)

very rarely used instructions -originally present to allowtranslation of 8085 programs to 8086.

IN, OUT



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Examples:

IN AX, 0C8h ;reads port address at C8h (8 bit address) and loads into AX

IN AL, DX ;reads the port address given by DX and loads into AL

OUT p8 ,AX ;sends the data in AX to port p8

OUT DL, AX ; sends the data in AX to port given by DL

IN reads data from the specified IO port (8-bit or 16-bit wide) to the accumulator (

AL or AX).

The IO port can be an immediate address (8-bit only) or specified by a variable

or register (8 or 16-bit address). (Seems only DX can be used.)

OUT sends data from the accumulator register to the specified I/O port. Both byte andword sized data may be sent using IN and OUT.

II. Arithmetic Instructions(20)



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Intel 8088 has 20 instructions for performing integer addition,Subtraction , multiplication, division, and conversions from binary codeddecimal to binary.

Arithmetic Instructions (cont.)



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Addition



57

Binary addition of two bytes or two words are performed using:

ADD d,s

ADD adds bytes or words in d and s and stores result in d.The operands d and s can use the same addressing modes as in

MOV.

Addition of double-word is achieved by using the carry bit in the

FLAGS register. The instruction

ADC d,s

automatically includes the carry flag, and is used to add the

more significant word in a double-word addition.

Addition

Example: addition of two double words stored at [x] and [y]MOV AX, [x] ; Loads AX with the word stored at location [x]MOV DX, [x+2] ; Loads the high order wordADD AX, [y] ; Adds the low order word at [y]ADC DX, [y+2] ; Add including the carry from the low order words

Addition (cont.)



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Example: addition of two double words stored at [x] and [y]

Addition of binary coded decimal numbers (BCD) can be performedby using ADD or ADC followed by the DAA instruction to convert thenumber in register AL to a BCD representation. (see example)

Addition of numbers in their ASCII form is achieved by using AAA(ascii adjust after addition).

ASCII adjust for Addition (AAA)



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ASCII codes for the numbers 0 to 9 are 30H to 39H respectively. The ascii adjust instructions convert the sum stored in AL to two-byte unpackBCD number which are placed in AX.

When 30H is added to each byte, the result is the ASCII codes of the digitsrepresenting the decimal for the original number in AL.Example: Register AL contains 31H (the ASCII code for 1), BL contains 39H (theASCII code for 9).ADD AL, BL ; produces the result 6AH which is kept in AL.AAA ; converts 6AH in AL to 0100H in AX

Addition of 30H to each byte of AX produces the result 3130H (the ASCII codefor 10 which is the result of 1+9)

Subtraction



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Multiplication



61

Division



62

SIGN EXTENDED INSTRUCTIONS



63

Other Arithmetic Instructions



64

III. Logic and bit MANIPULATION Instructions (12)



65

III. Logic and bit MANIPULATION Instructions (12) (Contd)



66

Shift and Rotate



67

Shift and Rotate(Contd)



68

packed BCD from two ASCII characters



69

IV. Strings Instruction (6)



70

IV. Instruction for moving strings(Contd)



72

REP prefix



73

LODS and STOS string instructions



74

LODS and STOS string instructions(Contd)



75

CMPS and SCAS string instructions



76

V. Program Flow Instruction



77

Program Flow Instruction



78

Unconditional JUMP



79

Unconditional JUMP (cont.)



80

List file of program Demonstrating “backward” JMP



81

List file of program demonstrating “forward” JMP



82

Conditional Jumps



83

8086 Conditional Jump Instructions



84

Ex.: Reading ASCII code when a strobe is present



85

Assembly language for Reading ASCII code when a strobe is present



86

Loops



87

Procedures and modular Programming



88



89



90

Procedures and modular Programming (Contd)



91

Procedures



92

Stack Diagram



93

Using PUSH and POP instructions



94

Interrupt Instructions



95

Interrupt Instructions



96



97

Program Data and Storage



98

• Assembler directives for data storage

– DB - byte(s) – DW - word(s)

– DD - doubleword(s)

– DQ - quadword(s) – DT - tenbyte(s)

Arrays



99

• Any consecutive storage locations of the

same size can be called an array

X DW 40CH,10B,-13,0

Y DB 'This is an array'

Z DD -109236, FFFFFFFFH, -1, 100B

• Components of X are at X, X+2, X+4, X+8

• Components of Y are at Y, Y+1, …, Y+15

• Components of Z are at Z, Z+4, Z+8, Z+12

DUPll f l i b



100

• Allows a sequence of storage locations to be

defined or reserved

• Only used as an operand of a define

directive

DB 40 DUP (?)

DW 10h DUP (0)

DB 3 dup ("ABC")

db 4 dup(3 dup (0,1), 2 dup('$'))

Word Storage• Word doubleword and quadword data are



101

• Word, doubleword, and quadword data are

stored in reverse byte order (in memory)

Directive Bytes in Storage

DW 256 00 01

DD 1234567H 67 45 23 01

DQ 10 0A 00 00 00 00 00 00 00X DW 35DAh DA 35

Low byte of X is at X, high byte of X is at X+1

EQU Directive• name EQU expression



102

• name EQU expression

– expression can be string or numeric

– Use < and > to specify a string EQU

– these symbols cannot be redefined later in the

program

sample EQU 7FhaString EQU <1.234>

message EQU <This is a message>



103



104



105



106



107



108



109



110



111



112



113



114



115



116



117



118

Macro definition:



119

Macro definition:

name MACRO [parameters,...]

<instructions>

ENDMMyMacro MACRO p1, p2, p3

MOV AX, p1MOV BX, p2MOV CX, p3

ENDM

ORG 100h

MyMacro 1, 2, 3

MyMacro 4, 5, DX

RET

The syntax for procedure declaration:



120

The syntax for procedure declaration:name PROC

; here goes the code; of the procedure ...

RET

name ENDP

ORG 100h



121

ORG 100h

CALL m1

MOV AX, 2

RET ; return to Main Program.

m1 PROCMOV BX, 5RET ; return to caller.

m1 ENDP

END

ORG 100h



122

MOV AL, 1MOV BL, 2

CALL m2CALL m2CALL m2CALL m2

RET ; return to operating system.

m2 PROCMUL BL ; AX = AL * BL.

RET ; return to caller.m2 ENDP

END

addressing modes&instructions

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