8086 Software Programs

LAB PROGRAMS (SAI VIDYA INSTITUTE OF TECHNOLOGY)A.

Prof. A.M. Padma Reddy (HOD), SVIT, CSE Dept, Bangalore Page 1

1

MICROPROCESSOR LAB

MANUAL (SOFTWARE

PROGRAMS)

Note: Only programs have been given. The method of writing the programs and

designing of 8086 program using algorithms is explained in the text book:

“Introduction to microprocessors – 8086”

by: Prof. Padma Reddy & Prof. Y. Jayasimha

A. LAB PROGRAMS (SAI VIDYA INSTITUTE OF TECHNOLOGY)


2

AAPPPPEENNDDIIXX -- LLAABB PPRROOGGRRAAMMSS

Problem 1.(a): Search a key element in a list of 'n' 16-bit numbers using the binary

search algorithm (Detailed design refer section 8.10.3, page 8.77)

; Program Name: BINARY SEARCH (BSRCH.ASM)

; Author : A.M. Padma Reddy and YJ ; Purpose : To search for the key in a given list using binary search

; Inputs : List of elements, number of elements, key

; Outputs : Success if key found, Unsuccessful if key not found

.MODEL SMALL

; Macro definitions

; Macro definition to display the message

PRINTF MACRO MSG ; Here, MSG is the parameter to the macro

MOV AH, 09H

MOV DX, OFFSET MSG

INT 21H

ENDM

; Macro to terminate the program (return control to DOS)

EXIT MACRO

MOV AH, 4CH

INT 21H

ENDM

;---------------------------Data segment starts------------------------------

.DATA

A DW 1111H, 2222H, 3333H, 4444H, 5555H

N DW ($-A)/2 ; Number of terms of N

KEY DW 5555H ; Key to be searched

LOW_ DW ?

HIGH_ DW ?

MID DW ?

MSG1 DB “SUCCESSFUL SEARCH”, 0AH, 0DH, 24H

MSG2 DB “UNSUCCESSFUL SEARCH”, 0AH, 0DH, 24H

;---------------------------Data segment ends-------------------------------



3

;---------------------------Code segment starts------------------------------

.CODE

begin: ; Initialization of data segment register

MOV AX, @DATA

MOV DS, AX

;Initialization for the loop

MOV LOW, 0 ; low ← 0

MOV AX, N ; AX← N

MOV HIGH_, AX

DEC HIGH_ ; high ← N -1

;Search using binary search

L1: MOV SI, LOW_

CMP SI, HIGH_

JG L4 ;if ( low > high ) L4

; mid (low + high ) / 2

ADD SI, HIGH_ ; SI ← low + high

SHR SI, 1 ; SI ← (low + high)/2

MOV MID, SI ; mid (low + high ) / 2

;if ( key != a[mid] ) L2

MOV AX, KEY

MOV SI, MID

SHL SI, 1 ; SI should be twice mid

CMP AX, A[SI]

JNE L2

PRINTF MSG1 ; print search success

EXIT ; Terminate program

L2: CMP AX, A[SI] ; if ( key > a[mid] ) L3

JG L3

; high mid – 1

MOV AX, MID ; AX mid

DEC AX ; AX mid – 1



4

MOV HIGH_, AX ; high mid – 1

JMP L1

; low mid + 1

L3: MOV AX, MID ; AX mid

INC AX ; AX mid + 1

MOV LOW_, AX ; low mid + 1

JMP L1

L4: PRINTF MSG2 ; print search unsuccessful

EXIT ; Terminate program

END begin

;---------------------------Code segment ends------------------------------

Note: Since LOW and HIGH are assembler directives in 8086 assembly language, we use

the identifiers LOW_ and HIGH_ instead of LOW and HIGH in the above program



5

Problem 2: Write ALP macros:

i) to read a character from the keyboard in the module(1) (in a different file)

ii) to display a character in module (2) (from different file)

iii) use the above two modules to read a string of characters from the keyboard

terminated by the carriage return and print the string on the display in the next line

Note: Type two programs and save using the names 2A.ASM and 2B.ASM as shown

below:

.model small

.stack

;---------------------Data segment starts--------------

.data

extrn a:byte ;Defines that an array variable a is external to the file

msg1 db 0ah,0dh,”Enter the characters:”,24h

msg2 db 0ah, 0dh,”The typed characters:”,24h

;--------------------- Data segment ends---------------

;Macro to display the message

printf macro msg

mov ah, 09h ; function to display the text

lea dx, msg ; offset value of text to be displayed

int 21h

endm

;macro to terminate the program

exit macro

mov ah, 4ch ; function to terminate the program

int 21h

endm

;------------------Code segment starts----------------

.code

extrn getchar:far ;Function getchar is a defined in another file

extrn putchar:far ;Function putchar is defined in another file

2A.ASM



6

begin: mov ax,@data

mov ds,ax ; Initialization of data segment register

printf msg1 ; printf(“Enter the characters”);

mov si, 00 ; Initialize the index si to 0

next: call getchar ; Read a character

cmp al, 0dh ; Is character read is end of line

je over

mov a[si], al ; Store the character in the array

inc si ; point to next location in array a

jmp next ; Read the next character

over: mov cx, si ; Numer of characters to be displayed

printf msg2 ; printf(“The typed characters are:”);

mov si, 00 ; Initialize si to point to beginning of a

up: mov al,a[si] ; Obtain the character from array

call putchar ; Display the current character

inc si ; Point si to next character

loop up

mov ah,4ch

int 21h

end begin

;end of the program

Note: After typing 2A.ASM, Give the following commands:

C:\>MASM 2A;

C:\>EDIT 2B.ASM

and type the program given in next page



7

.model small

.stack

; ---------------------Start of data segment -----------------------------------

.data

public a ;The data array is being made public allocating

a db 10H dup(?) ; Reserved 10 locations. external function can access a

;-------------------- End of data segment ------------------------------------

;--------------------Start of code segment----------------------

.code

public getchar ; External program can access getchar()

public putchar ; External program can access putchar()

;Procedure to read key is made public

getchar proc far

mov ah,01 ; Function to read a character

int 21h ; Character read is copied to AL register

ret

getchar endp

;Procedure to write a character

putchar proc far

mov ah,02 ; Function to display the character

mov dl,al ; Character to be displayed

int 21h ; Display the character

ret

putchar endp

end;

;---------------------End of code segment ------------------------

Note: After typing 2B.ASM, Give the following commands:

C:\>MASM 2b;

If no errors, type the following command

C:\>LINK 2A + 2B

C:> 2A Enter the characters: PADMAREDDY

The typed characters:PADMAREDDY

2B.ASM



8

Problem 3: Sort a given set of 'n' numbers in ascending and descending orders using the

bubble sort algorithm. (Refer section 8.10.4, page 8.81)

; Program Name: BUBBLE SORT (BSORT.ASM)

; Author : A.M. Padma Reddy and YJ ; Purpose : To sort the given elements using bubble sort

; Inputs : List of elements to sort

; Outputs : Sorted array

.MODEL SMALL

;---------------------------Data segment starts------------------------------

.DATA

A DW 5555H, 4444H, 3333H, 2222H, 1111H

N DW ($-A)/2 ; Number of terms of N

I DW ?

J DW ?

TEMP DW ?

;---------------------------Data segment ends-------------------------------


.CODE


MOV AX, @DATA

MOV DS, AX

MOV J, 1 ; Initial pass

MOV CX, N

DEC CX ; CX← N-1

L1: MOV SI, 0 ; Index to the first item

; of array A

MOV DX, N

SUB DX, J ; DX = N – J is the

; counter for inner loop

L2: MOV AX, A[SI] ; AX ← a[si]

CMP AX, A[SI+2] ; If a[si] <= a[si+2])



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JLE L3 ; goto L3

MOV TEMP, AX ; temp ← a[si]

MOV BX, A[SI+2] ; a[si] a[si+2]

MOV A[SI], BX

MOV A[SI+2], AX ; since AX contains

; temp

ADD SI, 2 ; Point to next word

L3: DEC DX ; decrement inner loop

JNZ L2 ; if (DX !=0) goto L2

LOOP L1 ; if (CX !=0) goto L1

; i.e., goto next pass

INT 3H ; Stop execution at this

; point

END begin

;---------------------------Code segment ends------------------------------



10

Problem 4: Read an alphanumeric character and display its equivalent ASCII code at the

center of the screen. (See section 9.3.1, page 9.12)

.MODEL SMALL

; Macro defintions

; Macro to read a character

GETCHAR MACRO

MOV AH, 01 ; Function to read a character

INT 21H ; AL has the character read

ENDM

; Macro display a character

PUTCHAR MACRO char

MOV AH, 02 ; Function to display a character

MOV DL, char ; Character to be displayed

INT 21H

ENDM


PRINTF MACRO MSG

MOV AH, 09H ; Function to display the text

LEA DX, MSG ; Offset value of text to be displayed

INT 21H

ENDM

;Macro to ternminat the program

EXIT MACRO

MOV AH, 4CH ; Function to terminate the program

INT 21H

ENDM

;--------------------------Data segment starts------------------------

.DATA

MSG1 DB "Enter a character",0ah, 0dh, 24h

MSG2 DB "The ASCII value: ",24h

;--------------------------Data segment ends--------------------------



11

;--------------------------Code segment starts------------------------

.CODE

begin: MOV AX, @DATA ; Initialization of data segment register

MOV DS, AX

PRINTF MSG1 ; DISPLAY "Enter a character"

GETCHAR ; Read a character from keyboard

MOV BH, AL ; Character in AL is copied to BL

MOV BL, AL ; Copy into BL register also

; Convert the lower nibble into ASCII

AND BL, 0FH ; Mask the higher nibble

CMP BL, 0AH

JL L1

ADD BL, 07H ; If greater than 9, ADD 7

L1: ADD BL, 30H ; Convert into ASCII by adding 30H

; Obtain ASCII value of higher nibble

AND BH, 0F0H ; Mask the lower nibble

MOV CL, 04

SHR BH, CL ; Move to higher nibble to lower nibble

CMP BH, 0AH

JL L2

ADD BH, 07H ; If greater than 9, add 7

L2: ADD BH, 30H ; Conver into ASCII by adding 30H

PRINTF MSG2 ; Display "The ASCII Value ="

PUTCHAR BH ; display higher nibble of ASCII value

PUTCHAR BL ; display lower nibble of ASCII value

EXIT

END begin



12

Problem 5: Reverse a string and check whether it is a palindrome or not (see section 8.9.4, page 8.62)

; Program Name: PALINDROME(PALI.ASM)

; Author : A.M. Padma Reddy and YJ ; Purpose : To check whether the string is palindrome

; Inputs : Give the string

; Outputs : Palindrome or not a palindrome

.MODEL SMALL

; Macro definitions

; Macro definition to display the message

PRINTF MACRO MSG ; Here, MSG is the parameter to the macro

MOV AH, 09H

MOV DX, OFFSET MSG

INT 21H

ENDM

; Macro to terminate the program (return control to DOS)

EXIT MACRO

MOV AH, 4CH

INT 21H

ENDM

;----------- Data segment starts------------------ ; Memory representation

.DATA

STR1 DB “LIRIL” ; Given string

N DW $ – STR1 ; Length of given string

STR2 DB 10 DUP(?) ; Storage space for reversed string

MSG1 DB “String is palindrome”, 0AH, 0DH, 24H

MSG2 DB “String in not a palindrome”, 0AH, 0DH, 24H

;------------ Data segment ends-----------------


.CODE


MOV AX, @DATA

MOV DS, AX



13

;Initialization for reversing

MOV SI, 0 ; STR1[SI] points to first character of STR1

MOV DI, 0 ; STR2[DI] point to first character of STR2

MOV CX, N ; where N = 5

ADD SI, CX

DEC SI

L1: MOV AL, STR1[SI] ; Copy STR1[SI] to

MOV STR2[DI], AL ; STR2[DI]

DEC SI ; Point to previous character

INC DI ; Point to next character

LOOP L1 ; Copy all characters of STR1

; Reversing of string is complete

; Check for the palindrome

; Initialization

MOV SI, 0 ; SI points to beginning of STR1

MOV CX, N ; No. of characters to compare

L2: MOV AL, STR1[SI] ; Compare STR1[SI] with STR2[DI]

CMP AL, STR2[SI]

JE L3

PRINTF MSG2 ; String is not a palindrome

EXIT ; Terminate the program

L3: INC SI ; Point to next character of STR1&2

LOOP L2 ; Compare N times

PRINTF MSG1 ; String is a palindrome


END begin

; ----------------------------Code segment ends --------------------------------



14

Problem 6 : Read two strings, store them in locations STR1 and STR2.check whether

they are equal or not and display appropriated messages. Also display the length of the

stored strings. (see section 9.3.2, page 9.14)

.MODEL SMALL

; Macro definitions

; Macro to read a character

SCANF MACRO BUFFER

MOV AH, 0AH ; Function to read a string

LEA DX, BUFFER ; Address of the buffer

INT 21H ; AL has the character read

ENDM


PRINTF MACRO MSG

MOV AH, 09H ; Function to display the text

LEA DX, MSG ; Offset value of text to be displayed

INT 21H

ENDM

;Macro to display a character

PUTCHAR MACRO CHAR

MOV AH, 02

MOV DL, CHAR

INT 21H

ENDM

;Macro to terminate the program

EXIT MACRO

MOV AH, 4CH ; Function to terminate the program

INT 21H

ENDM

;--------------------------Data segment starts------------------------

.DATA

MSG1 DB 0AH, 0DH,"Enter the first string",0ah, 0dh, 24h

MSG2 DB 0AH,0DH,"Enter the second string”, 0ah, 0dh,24h

MSG3 DB 0AH,0DH“The two strings are equal”, 0AH, 0DH, 24H



15

MSG4 DB 0AH,0DH“The two strings are not equal”, 0AH, 0DH, 24H

MSG5 DB 0AH,0DH“No. of characters in first string :”, 24H

MSG6 DB 0AH, 0DH, “No. of characters in second string :”, 24H

BUFFER1 DB 10

N1 DB ?

A1 DB 10 DUP(?)

BUFFER2 DB 10

N2 DB ?

A2 DB 10 DUP(?)

;--------------------------Data segment ends--------------------------

;--------------------------Code segment starts------------------------

.CODE

begin: MOV AX, @DATA ; Initialization of data segment register

MOV DS, AX

MOV ES, AX ; Intialization of extra segment register

; It is must for string manipulation

PRINTF MSG1 ; printf(“Enter the first string”);

SCANF BUFFER1 ; scanf(“%s”,a1);

PRINTF MSG2 ; printf(“Enter the second string”);

SCANF BUFFER2 ; scanf(“%s”,a2);

; Check length of two strings

MOV AL, N1

CMP AL, N2

JE L1 ; If lengths are equal compare strings

PRINTF MSG4 ; printf(“Two strings are not equal”);

JMP L3

; Initialization before comparing

L1: LEA SI, A1 ; Offset value of first string

LEA DI, A2 ; Offset value of second string

MOV CL, N1 ; No of characters to be compared

CLD ; To increment increment index registers

REPE CMPSB ; Compare as long as they are equal



16

JCXZ L2 ; If CX is zero, two strings are equal

PRINTF MSG4 ; printf(“Two strings are not equal”);

JMP L3

L2: PRINTF MSG3 ; printf(“Two strings are equal”);

L3: PRINTF MSG5 ; printf(“No. of characters in first string:”);

ADD N1, 30H ; Convert to ASCII by adding 30h

PUTCHAR N1 ; display no of characters of first string

PRINTF MSG6 ; printf(“No. of charas in second string:”);

ADD N2, 30H ; Convert to ASCII by adding 30h

PUTCHAR N2 ; display no. of characters of second string


END begin

;-------------------------------End of code segment----------------------------------------------



17

Problem 7 : Read your name from the keyboard and display it at a specified location on

the screen in front of the message WHAT IS YOUR NAME? you must clear the entire

screen before display.

.model small

;Macro to clear the screen

clrscr macro

mov bh,07 ;Attribute of character

mov ah,06 ;Activity is to scroll

mov al,0 ;No. of lines to scroll-0 blank screen

mov ch,00 ;y-coordinate of upper left corner

mov cl,00 ;x-coordinate of upper right corner

mov dh,24 ;Number of rows; Lower right row

mov dl,80 ;Number of columns; Lower right column

int 10h ; clear the screen

endm

;Macro to set the cursor

;

setcursor macro row, col ; row and col are the parameters

mov ah, 02 ; Function to set the cursor

mov bh, 0 ; Current page

mov dh, row ; Which row

mov dl, col ; Which column

int 10h ; Set the cursor

endm

;Macro equvivalent to

; printf()

; scanf()

pri_scn macro val, msg ; val – function to read or display

; msg is the parameter to be displayed

mov dx, offset msg ; DX holds offset of message to be displayed

mov ah, 09 ; Function to display the message

int 21h ; Display the message

endm



18

;Macro terminate the program

;Return control to dos

;

stop macro

mov ah, 4ch ; Function to terminate

int 21h ; Terminate the program

endm

;

;Macro display the character

; putchar() function of C

;

putchar macro val ; val is the parameter to be displayed

mov dl,val ; Character to be displayed

mov ah,02 ; Function to write a character


endm

;-------------Start of the data segment---------

.data

name1 db "What is your name?",24h

msg db "Enter the name",0ah,0dh,24h

buffer db 10 ; Max. no. of charcters in the name

n db ? ; Space to store length of name

a db 20 dup(?) ; Space to store the name

row db 10 ; Display at this row

col db 20 ; Display at this column

;---------------Start of code segment -----------

.code

mov ax, @data

mov ds, ax

prn_scn 09, msg ;printf("Enter the numbers");

prn_scn 0ah, buffer ;scanf("%s",a);

clrscr ; Clear the screen



19

setcursor row, col ; Set the cursor

prn_scn 09, name1 ; printf(“What is your name?”);

mov cl,n

mov ch,0 ; CX: no. of characters to displayed

mov si,00 ; initial offset value w.r.t. array a

top: putchar a[si] ; Display the character

inc si ; Point to next character

loop top ; If not over go to top

stop ; Terminate the program

end begin



20

Problem 8 : Compute nCr using recursive procedure. Assume that 'n' and 'r' are non-

negative integers (See section 9.5.1.2, page 9.32 for details)

.MODEL SMALL

.STACK 64

;------------------------------Start of data segment--------------------------------

.DATA

N DB 6

K DB 3

RES DB 0

;------------------------------Start of data segment--------------------------------

;------------------------------Start of code segment--------------------------------

.CODE

NCK PROC

CMP BL, 0

JNE L1 ; if k = 0

ADD RES, 1 ; return 1;

RET

L1: CMP BL, AL

JNE L2 ; if k = n

ADD RES, 1 ; return 1;

RET

L2: CMP BL, 1

JNE L3 ; if k = 1

ADD RES, AL ; return n

RET

L3: DEC AL ; n = n -1

CMP BL, AL

JNE L4 ; if k = n

INC AL ; return n+1

ADD RES, AL

RET



21

L4: PUSH AX ; Compute n-1

Ck

PUSH BX

CALL NCK

POP BX

POP AX

DEC BX ; k = k -1

PUSH AX ; Compute n-1

Ck-1

PUSH BX

CALL NCK

POP BX

POP AX

RET

NCK ENDP

begin: MOV AX,@DATA ; Initialization of DS register

MOV DS,AX

MOV AL,N ;Compute nCk

MOV BL,K

CALL NCK

INT 3H ; Terminate the program

END begin

;-------------------------------end of the code segment ------------------------------------------

Output: Give the following commands

C:\> MASM NCK;

C:\> LINK NCK;

C:\> DEBUG NCK.EXE

-G

-D DS:0000

The following memory dump is obtained:

14D4:0000 D8 A0 0C 00 8A 1E 0D 00-E8 B5 FF CC 06 03 14 8F

n k nCk = 14H

6C3 = (20)10

= (14)16



22

Problem 9 : Read the current time from the system and display it in the standard format

on the screen.

.model small

.stack

;Macro display the character

putchar macro val

push ax

mov dl,val ; Character to display

mov ah,02 ; Function to display the character


pop ax

endm

;---------------Start of the code segment -----------------

.code

; Procedure to convert from HEX to ASCII and display

hex_asc proc

mov ah,00

mov bl,0ah ; Divide by 10 to obtain unpacked BCD

div bl ; Obtain the unpacked BCD in AH and AL

; Higher nibble in AL

; Lower nibble in AH

add ax,3030h ; Convert to ASCII value

putchar al ; Display the character in AL

putchar ah ; Display the character in AH

ret

disp endp

begin: mov ah, 2ch ; Function to read the system time

int 21h ; Read the system time

: CH:CL contains hours:minutes

; DH:DL contains second:millisec



23

mov al, ch ; Display the hour

call disp

putchar ':'

mov al,cl ; Display the minute

call disp

putchar ':'

mov al,dh ; Display the second

call disp

exit ; Terminate the program

end begin



24

Problem 10 : Program to simulate a decimal up-counter to display 00-99

.model small

.stack

;Macro to display a character

putchar macro val

push dx

push ax

mov dl,val ; Character to be displayed

mov ah,02 ; Function to read character

int 21h ; Read a character

pop ax

pop dx

endm

;Code segment starts

.code

;Procedure to convert packed BCD to ASCII and display the character

dec_asci proc

mov cl,04

mov bh,al ; AL has the packed bcd

shr bh,cl ; Move the higher nibble to lower nibble

mov bl,al

and bl,0fh ; Mask the higher nibble

add bx,3030h ; Convert to ASCII

putchar bh ; Display higher nibble

putchar bl ; Display lower nibble

putchar 0ah ; Move cursor to next line

putchar 0dh ; Move cursor to beginning of current line

ret

dec_asci endp



25

; Main program starts from here

begin: mov al,00

top: daa ; convert to decimal

call dec_asci ; convert to ASCII and display

cmp al,99h ; Have you reached 99

je down ; If yes go down

add al,01 ; Increment count by 1

jmp top ; perform earlier task

down:

mov ah,4ch ; Function to terminate the program


end begin



26

Problem 11 : read a pair of input co=ordinates in BCD and move the cursor to the

specified location on the screen.

.model small

; Cursor to set at the specified row and column

setcursor macro row, col

mov ah, 02 ; Function to set cursor

mov bh, 00 ; page number

mov dh, row ; Set at this row

mov dl, col ; Set at this column

int 10h

endm


exit macro

mov ah, 4ch ; Function to terminate the program


endm

;Macro to read a character

getchar macro

mov ah,01h ; Function to read a character

int 21h ; AL contains the character typed

endm


printf macro msg ; msg is the parameter to be displayed

mov ah, 09 ; Function to display the message

mov dx, offset msg ; DX hold offset of message to be displayed

int 21h ; Display the message

endm

.data

row db ? ; Space to store row value

col db ? ; Space to store column value

msg1 db "Enter the row value :",0ah,0dh,24h

msg2 db "Enter the col value :",0ah,0dh,24h

.code



27

; Procedure to read a byte and convert it to packed BCD

read proc

getchar ; Get an ASCII value

sub al, 30h ; Convert into unpacked BCD

mov cl, 4

shl al, cl ; Make it higher nibble

mov bl, al ; Save higher nibble in BL

getchar ; Get next ASCII value

sub al, 30h ; Convert into unpacked BCD

add al, bl ; Convert 2 unpacked BCD to packed BCD

ret

read endp

begin: mov ax, @data ; Initialization of DS register

mov ds, ax

printf msg1 ; printf("Enter row value");

call read ; scanf("%d",&row);

mov row, al ; Save the row value

printf msg2 ; printf("Enter col value");

call read ; scanf("%d",&col);

mov col, al ; Save the column value

setcursor row,col ; Set the cursor at specified row, column

getchar ; Read a character


end begin

C:\MASM615>cursor

Enter the row value :10

Enter the col value :20

Note: Observe on the screen that cursor will be at row 10 and column 20



28

Problem 12 : program to create a file (input file) and to delete an existing file

.model small

.stack

; Macro to display the message and read the string

prn_scn macro val,msg ; val = 09 display the message msg

; val = 09 read into msg

mov ah,val ; Function to read or print

mov dx,offset msg ; Offset value of message or buffer

int 21h ; Read or write

endm


exit macro

mov ah,4ch ; Function to terminate the program


endm

.data

msg1 db "Enter the file :",0ah,0dh,24h

buff db 20 ; Max. no. of characters in the file

len db ? ; Length of file name

fname db 20 dup(?) ; Space for storing the file name

msg2 db 0ah,0dh,"File successfully deleted",24h

msg3 db 0ah,0dh,"Error in deleting: File may not exist or readonly",24h

;--------------------Start of code segment -------------------

.code

mov ax,@data

mov ds,ax

prn_scn 09,msg1 ; printf(“Enter the file name”);;

prn_scn 0ah, buff; ; scanf(“%s”,fname);



29

;Attach ASCIIZ at the end of the file

mov bx,00

add bl,len ; Length of file name

mov fname[bx],0 ; Store ASCIIZ at the end of file name

;Delete the file

mov ah,41h ; Function to delete the file

mov dx,offset fname ; DX contains name of file

int 21h ; Delete the file

jc fail ; If Carray is set delete failed

prn_scn 09,msg2 ; File deleted successfully


fail: prn_scn 09h,msg3 ; File not deleted


end begin

Documents

8086 Software Programs