8086 instruction

7/29/2019 8086 instruction

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for the example slides do not clickthe mouse to see the full

Microprocessor 1

Dr.Raed Al-qadi

2009for the example slides do not click

the mouse to see the full


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Instruction set for 8088/8086

microprocessors

1- Data transfer instructions.

2-ALU instructions.

3-Shift and rotate instructions.4-Program control and subroutines

instructions.

5-Miscallenous instructions.



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Data transfer instructions.

In 8088/8086 the data transfer or data movementinstructions are :

1) MOV

2) IN/OUT

3) LEA

4) MOV OFFSET

5) XLAT

6) PUSH and POP

7) String instructions

We are going to investigate these instructions in details.



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MOV

MOV instrucion has the following syntax:

MOV dst,src ; just clone the src into the dst

The following criteria's should be appliedwhereever MOV is used :

1- dst and src must be the same size.

2- dst must NOT be immediate value.

3- dst and src cant be segment registers at thesame time.

4- dst and src cant be memory at the same timeexcept with MOVS instruction ( discussed later).


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Example 1 :

*Knowing that,

DS=2042H.

SI=500H,AX=214E,

MOV [SI],AX ;

DS=2042H DS=20420H

SI=500

20920H

20921H

21 4E

AH AL

4E

21


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Dos and Don'ts

Some special cases :

Case 1 : MOV CS,anything; DONT ( if you transferto CS problems occur since CS contains the

program code)Case 2 :MOV Segment, anything ; DO if anything

does not equal segment register

Case 3 : MOV segment,segemnt; DO NOT

Case 4: MOV mem,mem; DON'T, memory tomemory is not allowed except with MOVS


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IN/OUT

These two instructions are used to read from

and to a device.

IN instruction syntax:

IN AL,port address; (port


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IN is used to read from input device in separateI/O mode.

Example 2:

IN AL,KEYBOARD; true

But,

ADD AL,KEYBOARD; false

IN AL,60H; true, means read from

Device at address 60H I/O addresses64Kbyte

0000

FFFF

KEYBOARD


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Example 3 :

The serial port is at 03F8H(>255), read a

byte from the serial port.

Since 03F8H>255

MOV DX,03F8H

IN AL,DX

I/O addresses

64Kbyte

0000

FFFF

data03F8H

DX=03F8H

AL=data


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Example 4 :

Define the DISPLAY device at 2430H and

read a byte from it.

DISPLAY EQU 2430H

MOV DX,DISPLAY

IN AL,DX I/O addresses64Kbyte

0000

FFFF

DISPLAY2430

DX=03F8H

AL=data


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OUT instruction

Syntax:

OUT port address, AL; can be AX or AL

OUT DX,AL ; Just holds for DX

Example 5:

Send char A to the printer at 03A0H.

MOV DX,03A0H

MOV AL,A

OUT DX,AL


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LEA

LEA stands for load effective address, it

loads a 16 bit register with the offset

address of the data specified by the

operand.

Syntax :

LEA dst,src; dst stores the address of src


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Example 6:

Fill the array AR1 with *

.org 300

.data AR1 DB 20 DUP(?)

LEA SI,AR1

MOV CX,20;

LP : MOV [SI],* INC SI

LOOP LP

SI=add(AR1)

AR1

Cx=201918

17161514131211

109876543

21*

***

******

****

***

***


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MOV OFFSET

The same as LEA ,

Syntax :

MOV dst,OFFSET src; store the addressof src into dst.


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XLAT

XLAT called translate instruction, the

operation ofXLAT is changing the value of

AL with the memory pointed to by (AL+BX)

Mem[AL+BX]AL

AL=data

AL=dddd

BX=mmmm

datadddd+mmmm


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More clearly the XLAT instruction converts

the content of register AL, using a table. Its

beginning address is contained in registerBX. The content of register AL is interpreted

as a relative address in the table.

XLAT is the only instruction that adds 8 bitnumber to 16 bit number!

XLAT never Affects the flags..


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Example 7:

Suppose that a seven segment LED display

lookup table is stored in memory at

address TABLE, now XLAT can use this

table to translate the BCD number in ALto a seven segment code also in AL.

Let TABLE =1000H,DS=1000H,


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First , obtain the look up table.

gfedcbax

0

x

1

x

2

x

3

01111110000

00001101000

10110110100

10011111100

11011100010

11011011010

11111010110

00001111110

11111110001

11011111001

3FH0

06H1

5BH2

4FH3

66H4

6DH5

7DH6

27H7

7FH8

6FH9

Next see the code


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The code for the operation of converting is verysimple using XLAT

TABLE DB3FH,06H,5BH,4FH,66H,6DH,7DH,27H,7FH,6FH

MOV AL,SOME_BCD_VALUE

MOV BX,OFFSET TABLE

XLAT

For example take 06H(6 BCD) for the initialvalue of AL,after translation AL contains 7DH,see the figures in the next slide.


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AL=4FH

TABLE

DB 3FH,06H,5BH

DB4FH,66H,6DH DB 7DH,27H,7FH,6FH

MOV AL,06H

MOV BX,OFFSET TABLE

XLAT

3FH

06H

5BH

4FH

66H

6DH

7DH

27H

7FH

6FH

DS TABLE

BX=address (TABLE)=1000H

1000H

AL=06H

Temp=AL+BX=1006HAL=memory [temp]

AL=4FH, which is the

seven segment

code of 06H


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Example 8:

Write code to read the input key from port(60H) and output the value to a 7-segemntdisplay at 80H.

KEYS EQU 60H; DISPLAY EQU 80H;

TABLE DB3FH,06H,5BH,4FH,66H,6DH,7DH,27H,7FH,6FH

LEA BX,TABLE;

IN AL,KEYS; XLAT

OUT DISPLAY,AL


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Stack instructions

PUSH and POP

Both are important to store and retrieve datafrom the LIFO structure (stack),

PUSH instruction :

In 8088 and 8086 PUSH always transfer 2bytes of data to the stack.

It has the following syntax :

PUSH src;src could be any 2 byte register or memory

location or immediate value


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How does PUSH executes?

Simply when we push 2 bytes of data to thestack, the most significant byte is stored in

stack segment pointed to by SP-1, and the least

significant to location pointed by SP-2,see thefollowing example,

Example 9 :let AX=31F2H,SP=100H,SS=2000H;

CX=FFFFH ,BX=413AH


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SP=FE

PUSH AX

SS=2000H

SP=100

31 F2

AX31

F2


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Example 10:

Assume SP=100H,SS=2000H

AX=31F2H,BX=413AH,CX=FFFFH.

We will see the changes in the stack after

executing the following sentences :

PUSH AX

PUSH BX

POP CX

POP AX


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SP=FCHSP=FEH

SS=2000H

SP=100H

PUSH AX

PUSH BX

POP CX

31 F2

AH AL

41 3A

BH BL

31F2413A

CH CL

3A41FF FF


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Example 11:

Write code to reverse string using stackoperations.

STR DB HELLO,00H ;

REVSTR DB 256 DUP(?)

MOV CX,0; count number of bytes

LEA SI,STR;

LEA DI,REVSTR


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NEXT-CHAR :

MOV AL,[SI]

CMP AL,00 JE NEXT

PUSH AX ;only care about AL

INC CX; JMP NEXT-CHAR


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NEXT:

CMP CX,00; check if cx =0 or the string is null

JE DONE-REV

POP AX MOV [DI],AL

INC DI

DEC CX;

JMP NEXT

DONE-REV: MOV [DI],00; to terminate with null


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String Instructions

String instructions

MOVS LODS STOS

STOSB STOSWLODSB LODSWMOVSB MOVSW

B stands for byte, and W stands for word


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MOVS*MOVSB : how does it execute?

Copy byte at DS:[SI] to ES:[DI]. Update SI and DI.Note that MOVS instruction deals with SI in thedata segment and DI in the extra segment. Theextra segment cant be overridden here).

DS:[SI] - ES:[DI]

if DF = 0 then (DF is D flag)SI = SI + 1

DI = DI + 1

elseSI = SI - 1

DI = DI - 1


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Using REP instruction with string

instructions MOV CX,100

REP MOVSB

How does the above code executes?

IF (Dflag==0)

{

Memory[DI]----memory[SI]

SI++;DI++;

}

else(If D flag==1)

{

SI--; DI-;CX--

}

IF(CX=!=0) repeat the instruction, other wise then next one

CLD clears the D flag (D=0)

STD sets the Dflag to 1(D=1)

D=1 we decrement SI,DI

D=0 we increment DI,SI


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CLD

MOV CX,100REP MOVSB ; is same as:

MOV CX,100;

LP:MOV AL,[SI]

MOV ES:[DI],AL

INC SI

INC DI

LOOP LP


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MOVSW *MOVSW : how does it execute?

Copy word at DS:[SI] to ES:[DI]. Update SI and DI.The same as MOVSB except we move a word notabyte, see what happens to SI and DI.

DS:[SI] - ES:[DI]

if DF = 0 thenSI = SI + 2

DI = DI + 2

elseSI = SI - 2

DI = DI - 2


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Example 13:

Copy 500 words from physical address

0800:200 to physical address

4000:0300,use string instructions.

0800 is the segment address and the 200 is the offset by contrast

4000 is the segment address and 0300 is its offset

So, we begin by assigning the segments

addresses to DS and ES and the offsets

to SI and DI


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MOV DS,0800H

MOV ES,4000H

MOV SI,0200H

MOV DI,0300H

CLD MOV CX,500

REP MOVSW1000

bytes

1000bytes

DS=0800H

ES=4000H

SI=200

DI=300

D=0

D

A

T

A


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STOSB and STOSW

ForSTOSB Store byte in AL into ES:[DI]. Update DI.

ForSTOSW Store word in AX into ES:[DI]. Update DI.How STOSB executes?

ES:[DI] = AL

if DF = 0 thenDI = DI + 1

elseDI = DI - 1 And STOSW executes the same as STOSB but DI is

incremented or decremented by 2


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LODSB and LODSW

ForLODSB this is how it is done

AL = DS:[SI]

if D = 0 thenSI = SI + 1

else

SI = SI 1And LODSW is the same except that SI is

changed by 2


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Example 14:

Write program to clear DOS-screen.

MOV ES,E000; E000:800 is not the real addressof video memory but here is just to illustrate the

idea MOV DI,800

CLD

MOV CX,25*80; since the screen is 25*80

characters and it is done by assembler MOV AX,0020H; 00 for color and 20H for space

REP STOSW


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The old screen is character screen,and the characters that appear

on it are stored in the video memory location,for each character

2 bytes are used one for the ASCCI and the other for the color

HELLO

H

E

L

L

O

00 20H

COLOR CHAR

Black space


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for the example slides do not click


Review of data movement

instructions

MOV operand1,operand2Copy operand2 to operand1

The MOV instruction cannot:

set the value of the CS and IP registers.

copy value of one segment register to anothersegment register (should copy to generalregister first).

copy immediate value to segment register

(should copy to general register first).

Transfer instructions never affect the flags


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LEA and MOV OFFSET do the same

operation, they load the address of one

memory location to a register.

XLAT copies the memory location

addressed by [AL+BX] to AL.

PUSH and POP are used to store and to

read from the stack, both cant deal with 8bits register, just use 16 bits.


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String instructions are :MOVSB,MOVSW,STOSB,STOSW,LODSB,LODSW.

The are used for moving or stroing orloading a whole byte or a whole word at atime, they shrink the code to minimal sizeat most problems

Only MOVS can move data from memoryto memory locations

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8086 instruction