Question

Write a program using 8086 assembly Language (with proper comments) that accepts three different digits as input from the keyboard. Each digit is first converted to a binary equivalent. The binary values of these three digits are compared and the middle value is put in the AL register. This AL register is multiplied with each value of a byte array of size 6, which is stored in the memory. The result of the multiplication is stored in the same memory location. You may assume the byte array has the values 02h, 06h, 08h, 03h, 01h, 05h. Make suitable assumptions, if any.

07 Aug 2023
Answer :
Word Count : 768

here's an 8086 assembly language program that fulfills your requirements. The program takes three different digits as input, converts them to binary equivalents, compares the binary values, and then multiplies the middle value with each element of a byte array. I'll provide comments to explain each part of the code.

ORG 100h   ; Set the origin address

MOV AH, 0   ; Clear AH register

; Accept three digits as input
MOV DX, 0   ; Clear DX register (will be used to store the input)
MOV AH, 1   ; Set AH to read character input
INT 21h     ; Read the first digit
SUB DL, '0' ; Convert ASCII to binary
MOV BX, DX  ; Store the first digit in BX

MOV AH, 1   ; Set AH to read character input
INT 21h     ; Read the second digit
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__________ _____.
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