Thursday, December 24, 2020

Assembly Language Program To Find Even And Odd Numbers In 8086


This program takes user input as an array and then determines the number of even numbers and prints them. This program is written using 8086 assembly language using emu8086 software.




TITLE PUCHTAA 
.MODEL SMALL 
.STACK 100H 
.DATA 
    MSG1 DB 0AH, 0DH, "THIS PROGRAM ACCEPTS SIZE OF ARRAY AND ELEMENTS THEN PRINTS EVEN NUMBERS$" 
    MSG DB 0AH, 0DH, "ENTER A LENGHT OF ARRAY: $" 
    ARRAY DB 0AH, 0DH, "ENTER ELEMENTS OF ARRAY$" 
    EVEN DB 0AH, 0DH, "TOTAL EVEN NUMBERS ARE $" 
    CONTINUE DB 0AH, 0DH, "CONTINUE? [Y/N] $" 
    NOEVEN DB 0AH, 0DH, "NO EVEN NUMBERS IN ARRAY!!$" 
 
 
 
 
.CODE 
    MAIN: 
    MOV AX, @DATA 
    MOV DS, AX 
     
    LEA DX, MSG1 
    MOV AH, 09H 
    INT 21H 
    CONT:    
    LEA DX, MSG 
    MOV AH, 09H 
    INT 21H 
     
    MOV AH, 01H 
    INT 21H 
     
    SUB AL, 30H 
    MOV CL, AL 
     
    LEA DX, ARRAY 
    MOV AH, 09H 
    INT 21H 
     
    MOV CH, 00H 
     
    AGAIN: 
    MOV DL, ' ' 
    MOV AH, 02H 
    INT 21H 
     
    MOV AH, 01H 
    INT 21H 
     
    SUB AL, 30H 
    MOV AH, 00H 
    MOV BL, 02H 
    AAD 
    DIV BL 
    CMP AH, 00H 
    JE INCREASE 
    RETURN: 
     
    DEC CL 
    CMP CL, 00H 
    JG AGAIN 
     
    CMP CH, 00H 
    JE NOEVENS 
     
    LEA DX, EVEN 
    MOV AH, 09H 
    INT 21H 
     
    JMP NUM 
     
 NOEVENS: 
    LEA DX, NOEVEN 
    MOV AH, 09H 
    INT 21H 
    JMP QUEST 
     
    
 NUM: 
     
    MOV DL, ' ' 
    MOV AH, 02H 
    INT 21H 
    
    POP AX 
    MOV DL, AH 
    ADD DL, 30H 
    MOV AH, 02H 
    INT 21H 
    DEC CH 
    CMP CH, 00H 
    JNE NUM 
 QUEST:    
    LEA DX, CONTINUE 
    MOV AH, 09H 
    INT 21H 
     
    MOV AH, 01H 
    INT 21H 
    OR AL, 20H 
    CMP AL, 'y' 
    JE CONT 
     
    JMP EXIT 
     
     
     
     
     
     
    INCREASE: 
    MOV AH, AL 
    ADD AH, AL 
    PUSH AX 
    INC CH 
    JMP RETURN 
     
    EXIT: 
    MOV AH, 4CH 
    INT 21H 
    END MAIN 

Finding Even Numbers In 8086 Assembly


Tuesday, May 19, 2020

[Solved] Creating Nested Loops in ascending order in Assembly Language

The following code demonstrates nested loops in ascending order example in 8086 Assembly Language Programming. Related program can be found here




TITLE PUCHTAA  
.MODEL SMALL  
.STACK 100H  
.DATA  
  
  
.CODE  
    MAIN:  
        MOV AX, @DATA  
        MOV DS, AX  
          
        MOV CL, 31H  
    LOOP1:  
        MOV DL, CL  
    LOOP2:  
        MOV AH, 02H  
        INT 21H  
          
        INC DL  
        CMP DL, 37H  
        JNG LOOP2  
          
        MOV DL, 0AH  
        MOV AH, 02H  
        INT 21H  
          
          
        MOV DL, 0DH  
        MOV AH, 02H  
        INT 21H  
          
          
        INC CL  
        CMP CL, 37H  
        JNG LOOP1  
          
           
          
        MOV AH, 04CH  
        INT 21H  
    END MAIN 
 
OUTPUT OF THE PROGRAM 
 
NESTED LOOPS IN ASCENDING ORDER IN ASSEMBLY LANGUAGE PROGRAMMING
 

Saturday, May 16, 2020

How to Print Sum of Series in Assembly Language


This Program Prints the sum of a series, of a number which is input by the user.



TITLE PUCHTAA 
.MODEL SMALL 
.STACK 100H 
.DATA 
    INTRO DB 0AH, 0DH, 07H, 09H, "THIS PROGRAM PRINTS THE SUM OF SERIES$" 
    MSG DB 0AH, 0DH, "ENTER A NUMBER: $" 
    NEWLINE DB 0AH, 0DH, "$" 
    DOAGAIN DB 0AH, 0DH, "DO AGAIN? Y FOR YES ELSE FOR NO: $" 
.CODE 
 
    MAIN: 
        MOV AX, @DATA 
        MOV DS, AX 
 
        LEA DX, INTRO 
        MOV AH, 09H 
        INT 21H 
 
         
    NUMBER:     
        LEA DX, MSG 
        MOV AH, 09H 
        INT 21H 
         
        MOV AH, 01H 
        INT 21H 
         
        SUB AL, 30H 
        CMP AL, 09H 
        JG NUMBER 
         
         
        MOV BL, AL 
        MOV BH, AL 
        PUSH BX 
        CMP AL, 00H 
        JE RESULT 
   AGAIN: 
         
         
        DEC BL 
        ADD BH, BL 
        CMP BL, 00H 
        JG AGAIN 
         
        MOV AH, 00H 
        MOV AL, BH 
        MOV BL, 0AH 
        DIV BL 
        MOV BX, AX 
        POP CX 
  RESULT:       
        LEA DX, NEWLINE 
        MOV AH, 09H 
        INT 21H 
        MOV CH, 00H 
         
  DISPLAY: 
        MOV DL,CH 
        ADD DL, 30H 
        MOV AH, 02H 
        INT 21H 
         
        MOV DL, '+' 
        MOV AH, 02H 
        INT 21H 
         
        INC CH 
        CMP CH, CL 
        JLE DISPLAY  
         
         
        MOV DL, 08H 
        MOV AH, 02H 
        INT 21H 
         
        MOV DL, '=' 
        MOV AH, 02H 
        INT 21H 
         
         
         
        MOV DL,BL 
        ADD DL, 30H 
        MOV AH, 02H 
        INT 21H       
         
        MOV DL, BH 
        ADD DL, 30H 
        MOV AH, 02H 
        INT 21H 
         
        LEA DX, DOAGAIN 
        MOV AH, 09H 
        INT 21H 
         
        MOV AH, 01H 
        INT 21H 
        OR AL, 20H 
        CMP AL, 'y' 
        JE NUMBER 
         
         
        MOV AH, 4CH 
        INT 21H 
    END MAIN 

OUTPUT of the program

Sum of Series Programming in Assembly Language
  • assembly language program for sum of series
  • sum of series in assembly language
  • series addition azevedotechcrunch

Click to Download Code

Download

Thursday, March 26, 2020

How To Find Length Of String In 8086 Assembly Language [Click to Download Code]


This program prints the length of user input string in 8085/8086 microprocessor assembly language programming


TITLE PUCHTAA 
.MODEL SMALL 
.STACK 100H 
.DATA 
 
    MSG1 DB 0AH, 0DH, "ENTER A STRING--> $" 
    MSG2 DB 0AH, 0DH, "NUMBER OF CHARACTERS ENTERED--> $" 
    INMSG DB 25 DUP('0') 
     
.CODE 
    MAIN: 
        MOV AX, @DATA 
        MOV DS, AX 
;----------------DISPLAY THE MESSAGE------------------ 
   AGAIN:      
        LEA DX, MSG1 
        MOV AH, 09H 
        INT 21H 
;----------------------------------------------------- 
         
;-----------TAKE STRING INPUT------------------------         
        LEA SI, INMSG 
        MOV DX, SI 
        MOV AH, 0AH 
        INT 21H 
         
        MOV SI, 02H 
        CMP INMSG+SI, 0DH 
        JE EXIT 
;---------------------------------------------------         
        LEA DX, MSG2 
        MOV AH, 09H 
        INT 21H 
;----------SET SOURCE INDEX(SI) VALUE TO 1---------- 
;----------BECAUSE FIRST INDEX TELLS NUMBER OF BYTES READ--------         
        MOV SI,01H 
         
        MOV AL, INMSG+SI 
        MOV AH, 00H 
        MOV BL, 0AH ; 
        AAD         ; 
        DIV BL      ;BREAKING THE VALUE IF IT EXCEEDS 10 
;--------------------------------------------------------         
        MOV DX, AX 
        ADD DL, 30H 
        MOV AH, 02H 
        INT 21H 
         
        MOV DL, DH 
        ADD DL, 30H 
        MOV AH, 02H 
        INT 21H 
        JMP AGAIN 
   EXIT:      
         
        MOV AH, 04CH 
        INT 21H 
    END MAIN


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How to Count Lenght of a given string in assembly language- OUTPUT

Wednesday, March 25, 2020

String Manipulation in 8086 Assembly Language | The Easy Way


This program asks user to input any string and then asks number of times it will print is. Then, it break it character wise.
TITLE PUCHTAA 
.MODEL SMALL 
.STACK 100H 
.DATA 
 
    MSG DB 0AH, 0DH, "ENTER YOUR NAME: $" 
    NUMBER DB 0AH, 0DH, "ENTER A DIGITI: $" 
    INNAME DB 20 DUP("?") 
.CODE 
 
    START: 
        MOV AX, @DATA 
        MOV DS, AX 
         
;;------------------------------------------------------------         
         
        LEA SI, INNAME 
         
        LEA DX, MSG 
        MOV AH, 09H 
        INT 21H 
;;------------------------------------------------------------         
        MOV DX, SI 
        MOV AH, 0AH 
        INT 21H  
;;------------------------------------------------------------         
        LEA DX, NUMBER 
        MOV AH, 09H 
        INT 21H 
         
        MOV AH, 01H 
        INT 21H 
;;------------------------------------------------------------         
        SUB AL, 30H 
        MOV BL, AL 
                  
                              
        MOV SI, 02H 
;;------------------------------------------------------------          
    AGAIN: 
      
        MOV DL, 0AH 
        MOV AH, 02H 
        INT 21H 
         
        MOV DL, 0DH 
        MOV AH, 02H 
        INT 21H 
                
     
        XOR BH, BH 
        PUSH BX 
;;------------------------------------------------------------ 
   DAGAIN: 
          
        MOV DL, INNAME+SI 
        MOV AH, 02H 
        INT 21H 
        DEC BL 
        CMP BL, 00H 
         
        JNZ DAGAIN 
;;------------------------------------------------------------         
        POP BX 
        INC SI 
        CMP INNAME+SI, 0DH 
        JNZ AGAIN 
;;------------------------------------------------------------         
         
         
         
         
        MOV AH, 04CH 
        INT 21H 
         
    END START         


Breaking String Character Wise and Manipulation in 8086 Assembly Language

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