Assembly Code

Posted on by Hridyesh singh bisht

General format of writing the program.

  1. under .model define the model you will be using such as small ,medium or large
  2. under.data initialize the variables and constants
  3. under.code write the code for particular task
  4. To exit the file we use ,mov ah,4ch ,int 21h
  5. end our code with , end .

A.Code to accept a two digit number from keyboard

  1. mov ax,@data and move ds,ax is done to point data segment register to data segment
  2. mov ah,01h and int 21 h is done to take input of first digit(4 bits)
  3. cmp value of al,3Ah to see see if the value is a character or a number
  4. if it’s a number then carry flag becomes =1
  5. JC down 1 means , if carry is generated then jump to instruction down 1
  6. if carry flag =0, then subtract 07h as difference between ASCII value of A and 1 is 7.
  7. convert ASCII value of Operand , by using AND gate.
  8. then we ROR , to switch the value from unit’s place to ten’s place
  9. then store that value in a temporary variable , assume bl
  10. for second input we repeat the steps done from2 -7
  11. now al has unit’s place and bl has ten’s place we add the two to get the number
.model small
;tells the memory and code model

.data
n1 db 05h
;is used to start the initialized and constant values 

.code
mov ax,@data
mov ds,ax
;This part is used to point data segment register to data segment in the memory

mov ah,01h
int 21h
;This part is used to input a number ,an int 21 h is an interrupt it contains functionality 

CMP al,3Ah
;This part is used to compare value with 3A , as it is used to differentiate between character and a number.

JC Down1
;Jump if carry , if it is a number then CF=1, otherwise 0

SUB al,07h
;subtract from characters ,to make it in range to input

Down1: AND al,0Fh
;We and the ASCII value and convert it into hexadecimal value

ROR al,04h
;To Switch the unit digit and tens place, by rotating left 4 bits

mov bl,al
;This is used to store al value in bl ,temporary
 
mov ah,01h
int 21hCMP al,3Ah
JC Down2
SUB al,07h
Down2: AND al,0Fh
;al here has the the unit place of 4 bits and bl has tens place of 4 bits 

ADD bl,al
;hence now in bl we got the two digit number

mov ah,4ch
int 21h
;DOS interrupt int 21/4Ch is EXIT - TERMINATE WITH RETURN CODE, the content of al is used as the return code and the process is terminate

end
;Assembler needs END directive to end the file. END can be written without Main. just end the file with END.

B.Code to Display 8 bit number

  1. The Accepting two digit number can be copied from 1st code ,
  2. BL contains the two digit number , and is moved to CL
  3. now to print 8 bit double digit number , we first have to print ten’s place and then , bl ,0f0h to get tens place of the 8 bit number
  4. then ror are it , to put it in units place
  5. cmp bl,0Ah to check if its alphabet or a number
  6. if its a alphabet then Carry flag=0 , then add 07 h ,as difference between ASCII value of A and 1 is 7.
  7. if carry flag =1, it means it was a number .
  8. Now add 30 to make it ASCII value of the number
  9. Move bl into dl for display
  10. then display the tenth’s value
  11. now repeat the process for 2 ND number display ,except extracting the ten’s place extract unit’s place by, AND BL,0Fh


MODEL SMALL

.DATA		

.CODE	
MOV AX,@DATA	
MOV DS,AX

CALL ACCEPT	;CALL TO PROCEDURE-ACCEPT

CALL DISP	;CALL TO PROCEDURE- DISP

MOV AH,4CH
INT 21h

ACCEPT PROC NEAR	;TO ACCEPT A 8 BIT NUMBER

MOV AH,01h      ;to  accept keyword input
INT 21h	

CMP AL,3Ah	;check if input is numeric or an alphabet
JC DOWN1
SUB AL,07h	;if alphabet is entered
DOWN1:	AND AL,0Fh;if number is entered
ROR AL,04h	;putting to tens place
MOV BL,AL

MOV AH,01h      ;to  accept keyword input
INT 21h	 

CMP AL,3Ah	;check if input is numeric or an alphabet
JC DOWN2
SUB AL,07h	;if alphabet is entered
DOWN2:	AND AL,0Fh;if number is entered
ADD BL,AL	;now BL has the final input 

RET
ENDP		;end of procedure

DISP PROC NEAR	;to display 8 bit number

MOV BL,CL
AND BL,0F0h		;to get tens place of the 8 bit number
ROR BL,4h		;to put it in units place
CMP BL,0Ah		;to check if its an alphabet or a number
JC DOWN3
ADD BL,07h		;if alphabet
DOWN3:	ADD BL,30h;if number
MOV DL,BL
MOV AH,02h		;printing DL register
INT 21h

MOV BL,CL
AND BL,0Fh		;to get units place of the 8 bit number
CMP BL,0Ah		;to check if its an alphabet or a number
JC DOWN4
ADD BL,07h		;if alphabet
DOWN4:	ADD BL,30h;if number
MOV DL,BL
MOV AH,02h		;printing DL register
INT 21h

RET
ENDP

END

C.Menu driven program :

  1. We have a macro for printing the message’s
  2. Lea is used to load the first index of a string or an array
  3. mov ah,09h is used to print the string
  4. First we have to select the choice we want and store it in c
  5. CMP C,31h ,ASCII value of 1 = 31 .Hence he wanted to use addition procedure
  6. We call Accept procedure for accepting each digit , hence we call it twice
  7. ADD CL , N1 ; add the 2 numbers and thus being saved to CL register and then we display it.
  8. We repeat it for other methods as well
MESS MACRO MSG ;Macro to print a message on console ,Mess name and msg is para meter

MOV AH,09h
LEA DX,MSG
;Load effective address
INT 21h
;Prints a string stored in the address in the DX Register.

ENDM
;end the macro

.MODEL SMALL

.DATA
N1 DB 00h;to accept number
N2 DB 00h
C DB 00h
MSG1 DB 0Ah,0dh, "MENU:$";to print menu to the console
$ to NULL 

MSG2 DB 0Ah,0dh, "1)ADDITION :$"
;0Ah,0dh is to display data on the next line 

MSG3 DB 0Ah,0dh, "2)SUBTRACTION :$"
MSG4 DB 0Ah,0dh, "3)MULTIPLICATION :$"
MSG6 DB 0Ah,0dh, "ENTER 1ST NUMBER=:$"
MSG7 DB 0Ah,0dh, "ENTER 2ND NUMBER=:$"
MSG8 DB 0Ah,0dh, "ANSWER=:$"

.CODE
MOV AX,@DATA
MOV DS,AX

MESS MSG1;printing menu
MESS MSG2
MESS MSG3
MESS MSG4
MOV AH,01h
INT 21h       ;accepting choice
MOV C,AL      ;saving the choice to a variable
CMP C,31h    ;Ascii value of 1 = 31 
JNE STOP1    ;Jump if stop is choice is not addition 
MESS MSG6
CALL ACCEPT  ;to accept a number
MOV N1,CL
MESS MSG7
CALL ACCEPT  ;to accept a number
ADD CL , N1  ; add the 2 numbers and thus being saved to CL register
MOV CL,N1
MESS MSG8
CALL DISP ;displaying the answer
JMP STOP; to jump to end of code

STOP1:CMP C,32h; to compare to ‘2’
JNE STOP2
MESS MSG6
CALL ACCEPT ;to accept a number
MOV N1,CL
MESS MSG7
CALL ACCEPT ;to accept a number
SUB N1,CL ;to subtract and save answer to CL 
MOV CL,N1
MESS MSG8
CALL DISP ;to display answer
JMP STOP

STOP2:CMP C,33h ;to compare to ‘3’
JNE STOP3
MESS MSG6
CALL ACCEPT ;to accept a number
MOV N1,CL
MESS MSG7
CALL ACCEPT ;to accept a number
MOV N2,CL
MOV AL,N1
MOV BL,N2
MUL BL; multiplying AL and BL and saving the answer in Al register
MOV CL,AL
MESS MSG8
CALL DISP ;to display answer
JMP STOP

STOP3: JMP STOP

STOP:

MOV AH,4CH
INT 21h

ACCEPT PROC NEAR;to accept 8 bit numbers
MOV AH,01h
INT 21h;to  accept keyword input
CMP AL,3Ah;check if input is numeric or an alphabet
JC DOWN1
SUB AL,07h;if alphabet is entered
DOWN1;AND AL,0Fh;if number is entered
ROR AL,04h;putting to tens place
MOV BL,AL

MOV AH,01h
INT 21h;to  accept keyword input
CMP AL,3Ah;check if input is numeric or an alphabet
JC DOWN2
SUB AL,07h;if alphabet is entered
DOWN2;AND AL,0Fh;if number is entered
ADD BL,AL;now BL has the final input 

RET
ENDP;end of procedure

DISP PROC NEAR;to display 8 bit numbers
MOV BL,CL
AND BL,0F0h;to get tens place of the 8 bit number
ROR BL,4h;to put it in units place
CMP BL,0Ah;to check if its an alphabet or a number
JC DOWN3
ADD BL,07h;if alphabet
DOWN3:ADD BL,30h;if number
MOV DL,BL
MOV AH,02h;printing DL register
INT 21h

MOV BL,CL
AND BL,0Fh;to get units place of the 8 bit number
CMP BL,0Ah;to check if its an alphabet or a number
JC DOWN4
ADD BL,07h;if alphabet
DOWN4:ADD BL,30h;if number
MOV DL,BL
MOV AH,02h;printing DL register
INT 21h
RET
ENDP

END

D.Assembly Language Program to perform 32/16 bit Division.

for 16 bit/8 bit ,

  1. Dividend -AX
  2. Quotient-al
  3. remainder-ah

for 32bit/16 bit

  1. Divident -DX AX
  2. Quotient-AX
  3. remainder-DX
  1. We first accept 4*8 bits of the Divident
  2. then we accept 2*8 bits of divisor
  3. DX has the higher 16 bits and AX has the lower 16 bits(divident)
  4. BX has the 16 bits (divisor)
  5. DIV BX (DX AX/BX)
  6. MOV R1,DH and MOV R2,DL , putting remainder in R1(high) and R2(low)
  7. Display R1 and R2
  8. MOV CX,AX , putting quotient in CX
  9. Display CX

No flag is altered by the DIV instruction.

MESS MACRO MSG;Macro to print a message on console
MOV AH,09h
LEA DX,MSG
INT 21h
ENDM

.MODEL SMALL

.DATA
MSG1 DB 0Ah,0dh, "ENTER 32 BIT NUMBER:$";messages to be shown to prompt user or inform user
MSG2 DB 0Ah,0dh, "ENTER 16 BIT NUMBER:$"
MSG3 DB 0Ah,0dh, "QUOTIENT=$"
MSG4 DB 0Ah,0dh, "REMAINDER=$"
N11 DB 00h; to store first 8 bits of the 32-bit number (DH)
N12 DB 00h; to store second 8 bits of the 32-bit number (DL)
N21 DB 00h; to store third 8 bits of the 32-bit number (AH)
N22 DB 00h; to store the last 8 bits of the 32-bit number (AL)
D1 DB 00h; to store first 8 bits of the 16-bit number
d2 db 00h; to store second 8 bits of the 16-bit number
R1 DB 00h; to store first 8 bits of the 16-bit number(remainder)
R2 DB 00h; to store second 8 bits of the 16-bit number(remainder)



.CODE
MOV AX,@DATA
MOV DS,AX
MESS MSG1
CALL ACCEPT;accept first 8 bits of the 32 bit number

MOV N11,BL

CALL ACCEPT; accept second 8 bits of the 32 bit number

MOV N12,BL

CALL ACCEPT; accept third 8 bits of the 32 bit number

MOV N21,BL

CALL ACCEPT; accept last 8 bits of the 32 bit number

MOV N22,BL

MESS MSG2

CALL ACCEPT; accept first 8 bits of the 16 bit number

MOV D1,BL

CALL ACCEPT; accept last 8 bits of the 16 bit number

MOV D2,BL; note that 32 bit number is stored like
;DX has the higher 16 bits and AX has the lower 16 bits
MOV DH,N11
MOV DL,N12
MOV AH,N21
MOV AL,N22

MOV BH,D1; moving the 2-8 bit numbers to BX register
MOV BL,D2

DIV BX; DX AX/BX

MOV R1,DH; putting remainder in R1(high) and R2(low)
MOV R2,DL

MOV CX,AX; putting quotient in CX
MESS MSG3
CALL DISP; displaying quotient

MOV CH,R1
MOV CL,R2
MESS MSG4
CALL DISP; displaying remainder

MOV AH,4CH
INT 21h

ACCEPT PROC NEAR; to accept 8 bit numbers
MOV AH,01h
INT 21h; to accept keyword input
CMP AL,3Ah; check if input is numeric or an alphabet
JC DOWN1
SUB AL,07h; if alphabet is entered
DOWN1: AND AL,0Fh; if number is entered
ROR AL,04h; putting to tens place
MOV BL,AL

MOV AH,01h
INT 21h; to accept keyword input
CMP AL,3Ah; check if input is numeric or an alphabet
JC DOWN2
SUB AL,07h; if alphabet is entered
DOWN2: AND AL,0Fh; if number is entered
ADD BL,AL; now BL has the final input 

RET
ENDP; end of procedure

DISP PROC NEAR; to display 16-bit numbers
MOV BH,CH
AND BH,0F0h; to get tens place of the 8 bit number
ROR BH,4h; to put it in units place
CMP BH,0Ah; to check if its an alphabet or a number
JC DOWN3
ADD BH,07h; if alphabet
DOWN3: ADD BH,30h; if number
MOV DL,BH
MOV AH,02h; printing DL register
INT 21h

MOV BH,CH
AND BH,0Fh; to get units place of the 8 bit number
CMP BH,0Ah; to check if its an alphabet or a number
JC DOWN4
ADD BH,07h; if alphabet
DOWN4: ADD BH,30h; if number
MOV DL,BH
MOV AH,02h; printing DL register
INT 21h

MOV BH,CH
AND BH,0F0h; to get tens place of the 8-bit number
ROR BH,4h; to put it in units place
CMP BH,0Ah; to check if its an alphabet or a number
JC DOWN5
ADD BH,07h; if alphabet
DOWN5: ADD BH,30h; if number
MOV DL,BH
MOV AH,02h; printing DL register
INT 21h

MOV BH,CH
AND BH,0Fh; to get units place of the 8 bit number
CMP BH,0Ah; to check if its an alphabet or a number
JC DOWN6
ADD BH,07h; if alphabet
DOWN6: ADD BH,30h; if number
MOV DL,BH
MOV AH,02h; printing DL register
INT 21h

RET
ENDP

END

E.String handling

  1. first index tells the actual max length initialized
  2. second index tells the actual length of the word
  3. we point si to starting of the string Str1
  4. we then accept string from console ,
  5. then we request the second value from string to check the length,
  6. then we add 30 to increment the value and make it hexadecimal
MESS MACRO MSG; to let us to print a string
MOV AH,09h
LEA DX,MSG
INT 21h
ENDM

.MODEL SMALL

.STACK 1000H

.DATA
STR1 DB 30,?,30 dup('$');blank string
;arr db max_length, actual length, max length dup(initialization)
;first char in string is max length , second is actual length , further on is the ASCII values of characters 

MSG1 DB 0Ah,0Dh,"ENTER STRING $";prompt message
MSG2 DB 0Ah,0Dh,"LENGTH IS : $"

.CODE
MOV AX,@DATA;start of code
MOV DS,AX


MESS MSG1

LEA SI, STR1; si will point to the start index of s

MOV AH,01Ah;to accept string from console
MOV DX,SI
INT 21h

MOV AL,[SI+1];moving the value in the second index (real length)

ADD AL,30h;to convert number to hexadecimal

MESS MSG2

MOV AH,02h;to print length
MOV DL,AL
INT 21h

MOV AH,4CH
INT 21h
END

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Published by Hridyesh singh bisht

Hridyesh Bisht is a Technical Knowledge Specialist at Suse, where he focuses on creating technical documentation, designing training courses, and building GitHub Actions to automate processes. He also analyzes and visualizes user feedback and metrics to enhance documentation and training materials. Hridyesh is actively involved in open-source projects, including contributing to AsyncAPI under the Google Season of Docs 2023 program. He developed detailed documentation and training content and delivered a talk at the AsyncAPI conference. Hridyesh is currently contributing to SigNoz(An open source data monitoring tool). Hridyesh believes that "Engineering is the closest thing to magic in the world." He thrives on contributing to technical communities as a developer, technical writer, and speaker, sharing knowledge and fostering a community. A former AWS Community Builder for nearly three years, he also organizes meetups for the Doon Technical Community to connect and empower local tech enthusiasts.