how to work in 8086 -...

1

DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING

Year & Sem : II & III

Name of the Subject: MC-9238 Microprocessor Lab Branch : MCA

HOW TO WORK IN 8086:

Press asm twice.

Type sg 0000

Press enter

Type da starting address (4000)

Press enter

Press new(n)

Give one space and type the mnemonic.

Press enter once and new(n) twice.

After completing the program, press shift + 1 (!)

To give input value: press exam byte(e) for 8 bit data. Give the location, type the data and

press next.

Press exam word (w) fro 16 bit data. Give the location, type the data and press next.

To execute the program : press go(g).

Give the starting address of the main program.

Press enter.

To view the result : press exam byte(e) for 8 bit data. Enter the location of the output.

2

Press exam word (w) for 16 bit data. Enter the location of the output.

Press e and r, to view the value stored in the register.

To display the program : press asm and type di space starting address space ending

address. Press enter.

To edit the program : press asm and give the corresponding address of the mnemonics to be

modified. Type the mnemonic and press enter.

8086 PROGRAMS

ARITHMETIC OPERATION

ADDITION PROGRAM:

MOV CX, 0000

MOV AX, [4200]

MOV BX, [4202]

ADD AX, BX

JNC LI

INC CX

LI: MOV [4206], CX

MOV [4204], AX

INT 0003

3

SUBTRACTION PROGRAM:

MOV CX, 0000

MOV AX, [4200]

MOV BX, [4202]

SUB AX, BX

JNC LI

INC CX

LI: MOV [4206], CX

MOV [4204], AX

INT 0003

DIVISON PROGRAM:

MOV AX, [4200]

MOV BX, [4202]

DIV BX

MOV [4206], AX

MOV AX, DX

MOV [4208], AX

INT 0003

4

MULTIPLICATION PROGRAM:

MOV AX, [4200]

MOV BX, [4202]

MUL BX

MOV [4206], AX

MOV AX, DX

MOV [4208], AX

INT 0003

LOGICAL OPERATION:

AND

MOV AX, [4100]

MOV BX, [4102]

AND AX, BX

MOV [4500], AX

INT 0003

OR

MOV AX, [4100]

MOV BX, [4102]

OR AX, BX

MOV [4500], AX

INT 0003

5

NOT

MOV AX, [4100]

NOT AX

MOV [4500], AX

INT 0003

SUBROUTINE - PROGRAM:

MOV AX, 3231

CALL ADDITION

PUSH AX

MOV CX, AX

INT 3

ADDITION: ADD AX, 4135

PUSH AX

MOV BX, 5314

POP AX

ADD AX, BX

RET

Observation:

Output: Register A

6

PROGRAM TO ACCESS STACK:

MOV AX, 3231

ADD AX, 4135

PUSH AX

MOV BX, 5314

POP AX

ADD AX, BX

MOV CX, BX

PUSH CX

INT 3

Observation: Output: Register A

PROGRAM TO FIND THE FACTORIAL OF A NUMBER:

MOV SI, 5000

MOV DI, 7000

MOV CX, [SI]

MOV AX, CX

DEC CX

AGAIN: MUL CX

DEC CX

JNZ AGAIN

MOV [DI], AX

INT 3

Observation: Input: at 5000 Output : at 7000

7

PROGRAM TO FIND SQUARE ROOT OF A PROGRAM:

MOV SI, 5000

MOV DI, 7000

MOV CX, 0001

MOV BX, 0000

MOV AL, [SI]

UP: SUB AL, CL

JL STORE

INC BL

ADD CL, 02

JMP UP

STORE: MOV [DI], BL

INT 3

Observation:

Input : at 5000

Output : at 7000

8

PROGRAM TO SORT THE NUMBERS IN DESCENDING ORDER:

MOV AH, O8H

DEC AH

LOOP3: MOC CL, AH

MOV BX, 4100H

LOOP2: MOV AL, [BX]

INC BX

CMP AL, [BX]

JNC LOOP1

MOV DL, [BX]

MOV [BX], AL

DEC BX

MOV [BX], DL

INX BX

LOOP1: DEC CL

JNZ LOOP2

DEC AH

JNZ LOOP3

INT 3

Observation:

Input : From 4100 to 4107

Output : From 4100 to 4107

9

PROGRAM TO SORT THE NUMBERS IN ASCENDING ORDER:

MOV AH, O8H

DEC AH

LOOP3: MOC CL, AH

MOV BX, 4100H

LOOP2: MOV AL, [BX]

INC BX

CMP AL, [BX]

JC LOOP1

MOV DL, [BX]

MOV [BX], AL

DEC BX

MOV [BX], DL

INX BX

LOOP1: DEC CL

JNZ LOOP2

DEC AH

JNZ LOOP3

INT 3

Observation:

Input : From 4100 to 4107

Output : From 4100 to 4107

10

STRING OERATION (copy string):

MOV SI, 4200

MOV DI, 4300

MOV CX, 0004

CLD

L1: MOVSB

LOOP L1

INT 0003

Observation:

Input : 4200

Output : 4300

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SEARCHING A STRING OERATION:

MOV SI, 4200

MOV DI, 4300

MOV CX, 0006

CLD

MOV AL, 0008

L1: SCASB

LOOP L1

DEC DI

MOV BL, [DI]

MOV [SI], BL

INT 0003

Observation:

Input : 4200

Output : 4300

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HOW TO WORK IN 8085:

Press ASM.

Enter the starting address (8000).

Press next (,).

Type the mnemonic.

Press enter twice.

After completing the program, press esc.

To give input data : press memory(m).

Type the location. Enter data.

Press next for each data to save.

Press esc.

To execute the program : press go (g).

Enter the starting address of the main program.

Press enter.

To view the output: press memory (m).

Give the location of the result.

Press next to see the subsequent output.

To view the program : press disp(z).

Give the starting and the ending address of the program.

Press next.

To edit the program : Press Asm and enter the corresponding address of the mnemonic to

be modified.

13

TRAFFIC LIGHT DESIGN USING 8255:

START: MVI A, 80H ADDRESS:

OUT 33

CALL DELAY

MVI A, 07H

OUT 30

MVI A, 08H

OUT 31

CALL DELAY

MVI A, 0BH

OUT 30

MVI A, 04H

OUT 31

CALL DELAY

MVI A, 0DH

OUT 30

MVI A, 02H

OUT 31

CALL DELAY

MVI A, 0EH

OUT 30H

MVI A, 01H

OUT 31

CALL DELAY

N-TO S 07,08

E-TO-W 0B,04

W-TO-E 0D,02

S-T0-N 0E,01

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JMP START

DELAY: LXI D, FFFFH

LOOP : DCX D

MOV A, E

ORA D

JNZ LOOP

RET

Interfacing connection : Through P6.

STEPPER MOTOR INTERFACING USING 8255(8086 KIT):

MOV AL, 80H

MOV DX, 0FFC6

OUT DX, AL

START: MOV AL, 09H

MOV DX, OFFC4

OUT DX, AL

CALL DELAY

MOV AL, 0CH

MOV DX, 0FFC4

OUT DX, AL

CALL DELAY

MOV AL, 06H

MOV DX, 0FFC4

PORT A=0FFCOH

PORT B=0FFC2H

PORT C=0FFC4H

CONT.REG=0FFC6H

CLOCKWISE ANTI-CLOCKWISE

O9 O3

OC 06

06 0C

03 09

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OUT DX, AL

CALL DELAY

MOV AL, 03H

MOV DX, 0FFC4

OUT DX, AL

CALL DELAY

JMP START

DELAY: MOV DI, 02H

LOOP 1: MOV CX, 0FFFFH

LOOP 2: LOOP LOO2

DEC DI

JNZ LOOP 1

RET

Interfacing connection: Through P3.

8253 INTERFACING WITH 8085:

MVI A, 36H (Counter0, mode3) ADDRESS:

OUT 33

MVI A, 00H

OUT 30

MVI A, 3C

OUT 30

RST 1

COUNTER 0 : 30H

COUNTER1 : 31H

COUNTER2 : 32H

CONTROL REG : 33H

16

Interfacing Connection:

Through P6.

Output:

In TCON1: Pin 1 :: + ve

Pin 4 :: - ve

DAC INTERFACING WITH 8085(SAWTHOOTH WAVE FORM):

MVI A, 80H

OUT DBH

MVI A, 00

LOOP: INR A

OUT D8

JNZ LOOP

Interfacing connection:

Through P3.

OUTPUT:

In JP4, out1 :: + ve

Gnd :: Gnd

PORT A : D8H

PORT B : D9H

PORTC : DAH

Control Reg : DBH

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ADC PROGRAMMING:

;NIFC07A WITH 85MEL. ADC POLLED MODE

ORG E000H ADDRESS:

LXI H, CHANNEL

MOV A, M

STA TEMP

MVI H, 00

MOV L, A

SHLD FFF7H

MVI A, 87H

AD00: MVI A, 90H

OUT CTL_PORT

LDA TEMP

OUT PORTC

MVI A, 0FH

OUT CTL_PORT

LXI D, 0AH

CALL DELAY

MVI A, 0EH

OUT CTL_PORT

MVI A, 0CH

OUT CTL_PORT

AD01: IN PORTA

ANI 80H

CPI 80H

JNZ AD01

MVI A, 0DH

OUT CTL_PORT

IN PORTA

PORTA : D8H

PORT B : D9H

PORT C : DAH

CTRL_PORT : DBH

UPDDT : 07D1H

CLR_DSP : 14A9H

LBL5 : 0042H

TEMP : F202H

CHANNEL : E100H

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STA FFF9H

MVI A, C7H

STA FFC2H

CALL UPDDT

CALL DELAY

JMP AD00

DELAY: LXI D, 1FFFH

BLOOP: DCX D

MOV A, D

ORA E

JNZ BLOOP

RET

RST 1

Interfacing Connection:

Through P3.

Input:

At E100 : give channel number as 07 to choose channel 8.

Output:

In IC0809, Pin 13 :: Gnd

Pin 12 :: VRef

Pin 5 :: VAin

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DC MOTOR SPEED CONTROLLER USNG 8085:

MEMORY

ADDRESS

OPCODE PROGRAM

8100 21 00 82 LXI H,8200

8103 3E 36 MVI A,36H

8105 D3 0B OUT 0BH

8107 7E MOV A, M

8108 D3 08 OUT O8H

810A 23 INX H

810B 7E MOV A, M

810C D3 08 OUT 08H

810E 76 HLT

MODE OF OPERATION:

8200=20H

8201=20H

FAST

8200=AAH

8201=AAH

SLOW

20

8279 INTERFACING WITH 8085(KEYBOARD/DISPLAY):

MVI A, 00H (if left entry 00 ; if right entry means 10)

OUT 31

MVI A, 31H

OUT 31

L2: MVI A, D0H

OUT 31

CALL DELAY

MVI A, 90H

OUT 31

LXI H, 8800

MVI C, 08H

L1: MOV A, M

OUT 30

CALL DELAY

INX H

DCR C

JNZ L1

CALL DELAY

JMP L2

RST 1

DELAY: MVI B, FFH

HERE2: MVI D, FFH

HERE1: DCR D

21

JNZ HERE1

DCR B

JNZ HERE2

RET

Interfacing connection:

Through P6.

Input: From 8800 to 8807

22

MASM SOFTWARE PROGRAMS

FILE MANIPULATION

FILE CREATION PROGRAM:

code segment

Assume cs:code,ds:code

org 1000h

mov ax,data

mov ds,ax

mov ah,3ch

mov cx,0000h

mov dx,1200h

INT 21h

mov ah,4ch

INT 21h

code ends

data segment

org 1200h

db'file.asm'

data ends

end

FILE DELATION PROGRAM:

code segment

Assume cs:code,ds:code

org 1000h

mov ax,data

mov ds,dx

mopv ah,41h

mov dx,1200h

int 21h

code ends

data segment

org 1200h

db'v1.asm '

data ends

end

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FILE RENAME PROGRAM:

code segment

Assume cs:code,ds:data,es:data

org 1000h

mov ax,data

mov dx,ax

mov es,ax

mov dx,1300h

mov[di],1500

lea dx,oldname

lea di.newname

mov ah,56h

int 21h

mov ah,4ch

int 21h

code ends

data segment

org 1300h

oldname db'file.txt'

newname db'file1.txt'

data ends

end

FILE MANIPULATION (writing into a file):

code segment

Assume cs:code,ds:data

org 1000h

mov dx,data

mov ds,ax

mov ah,3dh

mov a1,01h

lea dx,filename

int 21h

mov s1,1500h

mov s1,1500h

mov[s1],ax

mov ah,40h

lea dx,buffer

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mov ex,0050h

mov bx,[s1]

int 21h

mov ah,3ch

int 21h

mov ah,4ch

int 21h

code ends

data segment

org 1300h

filename db'vi.txt'

org 1400h

buffer db'vimicrosystems' data ends

end

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TERMINATE AND STAY RESIDENT (TSR) PROGRAM:

display EQU xxxxh

Screen EQU 0600h

delay1 EQU 10

.code

main PROC

in ax, Screen ; get screen status

push ax ; save status

mov dx,184Fh ;lower right corner (24,79)

mov bh,7 ; normalattribute

int 10h ; callBIOS

mov ah,2 ; locate cursor at0,0

mov bh, 0x7920 ; ASCII write green asforeground

L2: out display, ax ; create a delay loop

mov cx,delay1;

L3: push cx

L3a: loop L3a

pop cx

loop l3

sub al,1 ;raise time 1 sec

jnz L2

pop ax

exit

main ENDP

END main

26

ASSEMBLY LANGUAGE PROGRAMS USING ‘C’

The program prompts the user to enter the name of the subdirectory to be created. This name is

stored in the variable newdir. DX is loaded with address of newdir. Thus DX points to the

subdirectory name. Then DOS function call 39H is invoked to create the subdirectory.

If the carry flag is 0, it means that an error occurred during the creation of the subdirectory. In

such a case, a suitable error message is displayed on the CRT. If the carry flag is 0, it means that

the creation of the subdirectory was successful. In such a case, an appropriate success message

is displayed on the CRT.

PROGRAM:

#include <dos.h>

union REGS inregs,outregs;

char newdir[10];

void main()

{

printf("enter directory name \n");

scanf("%s",newdir);

inregs.x.dx=(int) &newdir;

inregs.h.ah=0x39;

intdos(&inregs,&outregs);

if(outregs.x.cflag!=0)

printf("ERROR!!.directory %s not created\n",newdir);

else

printf("Directory %s created in current directory\n",newdir);

}

OUTPUT:

Enter directory name : Sha

Directory Sha created in current directory.

27

String Replace

data segment

str db 'heffo$'

find db 'f'

repl db 'l'

count db 05h

data ends

code segment

assume cs:code,ds:data

start: mov ax,data

mov ds,ax

mov es,ax

k:mov cl,count

mov al,find

mov di,offset str

cld

repne scasb

mov al,repl

jz loop1

mov dx,offset str

mov ah,09h

int 21h

jmp e

loop1:dec di

stosb

jmp k

mov dx,offset str

mov ah,09h

int 21h

e: mov ah,4ch

int 21h

code ends

end start

28

String Copy

data segment

source db 'a','b','c'

des db 'g','h','i'

count db 03h

data ends

code segment


start: mov ax,data

mov ds,ax

mov es,ax

mov si,offset source

mov di,offset des

mov cl,count

cld

rep movsb

hlt

code ends

end start

String Length

data segment

src db 'preethi'

data ends

code segment


start: mov ax,data

mov ds,ax

mov di,offset src

mov al,00h

a:mov bl,[si]

cmp bl,00h

je h

jmp b

b:inc al

inc si

jmp a

29

h:hlt

code ends

end start

FILE MANIPULATION – CREATE A FILE


data segment

fn db 'new.asm','$'

data ends

code segment

org 1000h

mov ax,data

mov dx,ax

mov dx,offset fn

mov cx,00h

mov ah,3ch

int 21h

mov ah,4ch

int 21h

code ends

end

30

FILE MANIPULATION – WRITE A FILE


data segment

dabk db '1','2','3','4'

fn db 'new.asm'

data ends

code segment

org 1000h

mov ax,data

mov ds,ax

mov dx,offset fn

mov ah,3ch

int 21h

mov bx,ax

mov cx,10h

mov dx,offset dabk

mov ah,40h

int 21h

mov ah,4ch

int 21h

code ends

end

31

LINEAR SEARCH:

Code segment


Mov bx,1100h

Mov si,1102h

Mov cx,[si]

Add si,02h

Mov di,3000h

L2: mov ax,[si]

Cmp [bx],ax

Jz L1

Add si,02h

Loop L2

Jmp L3

L1: mov ax,1111h

Jmp L4

L3: mov ax,0000h

L4: mov [di],ax

Mov ah,4ch

Int 21h

Code ends

End

32

OUTPUT:

-e 1100

: 1100 00.89

-e 1102

:1102 00.15 00.65 00.63 00.89 00.56

-g

Program terminated normally.

-e 3000

:3000 11.11

SORTING:

Code segment


CLC

L2: mov bx,0000h

Mov si,2001h

Mov cx,[si]

Dec cx

Add si,02h

L1: mov ax,[si]

Cmp ax,[si+02]

Jc L

Xchg ax,[si+02]

Mov [si],ax

Mov bx,0001h

33

L: add si,02h

Loop L1

Dec bx

Jz L2

Mov ah,4ch

Int 21h

Code ends

OUTPUT:

-e 2001

:2001 00.05 00.00

-e 2003 .05 .00 .04 .00 .03 .00 .02 .00 .01 .00

-g

Program terminated normally

-e 2003

:2003 01. 00. 02. 00. 03.

:2008 00. 04. 00. 05.

MATRIX MULTIPLICATION:

code segment

assume cs:code

start: mov si,1000h

mov di,2000h

mov bp,3000h

a: mov al,[si]

mov bl,[di]

mul bl

mov bh, al

mov al,[si+1]

mov bl,[di+2]

34

mul bl

add al, bh

mov [bp],al

inc di

inc bp

mov al,[si]

mov bl,[di]

mul bl

mov bh, al

mov al,[si+1]

mov bl,[di+2]

mul bl

add al,bh

mov [bp],al

inc bp

inc si

inc si

dec di

loop a

int 3h

code ends

end start

OUTPUT:

-u

-e ds : 1000

01 01 01 01

-e ds : 2000

01 01 01 01

-g = cs:0000 0049

-d ss:3000

02 02 02 02

35

TSR Program:

#include"dos.h"

#include<conio.h>

#include<stdio.h>

//interrupt declarations

void interrupt (*prevtimer)();

void interrupt mytimer();

void writechar(char ch,int rows, int cols, int attr);

//a far pointer that will decrease video memory

char far *scr;

int a,b;

//our real program goes from here

void main()

{

scr=(char far*)0xb8000000;

prevtimer=getvect(8);

setvect(8,mytimer);

keep (0,1000);

}

//timer function

void interrupt mytimer()

{

a=random(25);

b=random(80);

writechar(' ',a,b,0);

(*prevtimer)();

}

//function that writes picked up character

void writechar(char ch,int row, int col, int attr)

{

*(scr+row*160+col*2)=ch;

*(scr+row*160+col*2+1)=attr;

}

36

Study of BIOS and DOS function calls for keyboard & Display interfacing

KEY BOARD BIOS assume cs:code,ds:data

data segment

msg db "Welcome to keyboard program",0ah,"$"

data ends

code segment

start: mov ax,data

mov ds,ax

mov dx,offset msg

mov ah,09h

int 21h

wait: mov ah,01h

int 16h

jz wait

mov ah,00h

int 16h

mov ah,0eh

int 10h

cmp al,1bh

je exit

jmp wait

exit: mov ax,4c00h

int 21h

ret

code ends

end start

37

DISPLAY BIOS


data segment

msg db "AaddsfDSs",0

data ends

code segment

start: mov ax,data

mov ds,ax

mov bx,offset msg

next: mov al,[bx]

cmp al,0

je stop

mov ah,0eh

int 10h

inc bx

jmp next

stop: mov ax,4c00h

int 21h

code ends

end start

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