Order Now  +1 678 648 4277 

Adding positive integers using x86 assembly in Linux assignment help

The assignment deals with reading, converting, and adding positive integer numbers using x86 assembly in Linux. The program uses interrupt 80h to read inputs and print outputs. The program uses only 16-bit instructions, using 32bit instructions for I/O only. The input numbers are read as strings which are converted to 16-bit integers internally. The final result is printed by first converting the 16-bit result to a string and then printing the converted string characters. Below is this assignment's solution provided by our x86 assembly assignment helpers.

Processing Positive Integers in Linux 

Adding positive integers using x86 assembly in Linux assignment help

Solution section .data title DB "This program will add two positive numbers which are less than 100.", 0x0a lenTitle EQU $ - title prompt1 DB "Enter first number: " lenPrompt1 EQU $ - prompt1 prompt2 DB "Enter second number: " lenPrompt2 EQU $ - prompt2 resultMsg DB "Addition result: " lenResultMsg EQU $ - resultMsg LF DB 0xa num1 DW 0; a place to save conversion of the first number num2 DW 0; the place to save conversion of the second number digit DB 0; a place to save digit of the result section .bss string1 RESB 3; a place to save a string of first number (2 digits + LF) string2 RESB 3; a place to save a string of second number (2 digits + LF) section .text global _start _start: ; Prompt user for the first number mov eax, 4 mov ebx, 1 mov ecx, prompt1 mov edx, lenPrompt1 int 80h ;Read and store the user input, only reads at most 2 characters mov eax, 3 mov ebx, 0 mov ecx, string1 ; save in string1 mov edx, 3 ; read at most 3 chars int 80h ; Prompt user for the second number mov eax, 4 mov ebx, 1 mov ecx, prompt2 mov edx, lenPrompt2 int 80h ;Read and store the user input, only reads at most 2 characters mov eax, 3 mov ebx, 0 mov ecx, string2 ; save in string2 mov edx, 3 ; read at most 3 chars int 80h ; convert ASCII string of the first number to an integer mov esi, string1; point to start of string1 with si mov bx, 0; bx will hold the converted number, start in zero conv1: xor cx, cx ; set cx to zero mov cl, [esi] ; load a character in cl CMP cl, 0xa; see if we reached the end of the string (LF char) JE endConv1; if we reached the end, exit loop sub cl, '0'; subtract 0 from the character to convert it to an integer mov ax, 10; load 10 in ax for making a multiplication mul BX; multiply the old number in bx by 10 mov bx, ax; save multiplication result in bx add bx, cx ; add the digit we converted to the number inc esi; advance to next digit character in the string jump conv1; repeat to convert next digits endConv1: at this point, bx has the value of the first number mov [num1], bx ; save first number in variable ; convert ASCII string of the second number to an integer mov esi, string2; point to start of string2 with esi mov bx, 0; bx will hold a converted number, start in zero conv2: xor cx, cx ; set cx to zero mov cl, [esi] ; load a character in cl CMP cl, 0xa; see if we reached the end of the string (LF char) je endConv2; if we reached the end, exit loop sub cl, '0'; subtract 0 from the character to convert it to an integer mov ax, 10; load 10 in ax for making a multiplication mul BX; multiply the old number in bx by 10 mov bx, ax; save multiplication result in bx add bx, cx ; add the digit we converted to the number inc esi; advance to next digit character in the string jmp conv2; repeat to convert next digits endConv2: at this point, bx has the value of the second number mov [num2], bx ; save second number in variable ; make the addition with the converted numbers mov ax, [num1] ; load first number add ax, [num2]; add to the second number and save the result in ax ; convert addition result to an ASCII digit string mov bx, 10 ; load 10 in cx for making division push bx; save a 10 in the stack conv3: mov dx, 0 ; clear dx to make division div bx; divide the old number in ax by 10 push dx; save digit in the stack CMP ax, 0; if the division of a number by 10 is zero, end loop jne conv3; otherwise keep dividing it by 10 ; print the result message mov eax, 4 mov ebx, 1 mov ecx, resultMsg mov edx, lenResultMsg int 80h ; print converted result print: pop ax; load a digit from the stack cmp ax, 10; if we found the 10 we have popped all digits, end loop je exit add al, '0'; else, convert digit to ASCII mov [digit], al; save the digit in the variable for printing it ; print the digit mov eax, 4 mov ebx, 1 mov ecx, digit mov edx, 1 int 80h jmp print; repeat to print all digits exit: ; print a line feed mov eax, 4 mov ebx, 1 mov ecx, LF mov edx, 1 int 80h ; exit the program mov eax, 1 xor ebx, ebx int 80h