Question

Write an 8088/8086 assembly program that counts the length of a null terminated string that starts at location STR. Print the result on the screen.

Answer 1

Greetings!!

Code:Tool used emu8086

org 100h

.data

STR DB "WELCOME TO 8086 PROGRAMMING!!!$"    ;load any string of choice

LENGTH DB 10,13,"LENGTH OF THE STRING = $"

.code

    mov ax,@data

    mov ds,ax

    MOV CL,0H           ;Initialize register AL with 0 to hold the count of the string  

    LEA SI,STR           ;Loading the address of the string STR into SI

NEXT:    

    MOV BL,[SI]         ;Load one byte from the array to the register BL

    CMP BL,24H          ;Compare whether the end of string $ reached

    JZ END                   ;If the character read is $,then go to the END label         

    INC CL                   ;if the character is not $, then increment the content of register AL ie increment character count

    INC SI                     ;increment the index of the array for reading the next character

  JMP NEXT             ;Loop back to the label to repeat the steps

END:

    ;display the message LENGTH OF THE STRING =  

    MOV DX,OFFSET LENGTH

    MOV AH,09H

    INT 21H

    ;converting hex to decimal

    MOV AH,0             ;clear ah

    MOV AL,CL          ;load count into AL

    MOV BL,10            ;load 10 to BL for division

    DIV BL                   ;divide AL with BL

    XOR AX,3030H    ;add 30 to both the bytes of the result in AX to convert it to the ascii equivalent digits

    MOV BX,AX       ;save into BX

    MOV DL,BL          ;load MSB of the answer to DL for display

    MOV AH,02H      ;parameter for display character

    INT 21H                 ;printing the MSB

    MOV DL,BH          ;load LSB to DL for display

    MOV AH,02H

    INT 21H                 ;display LSB

    MOV AX,4C00H

    INT 21H         

loop next

ret

Output screenshots:

Hope this helps