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Create GCD Routine In X86 Assembly Language, Recursive Algorithm Assignment Solution.


Write an x86 assignment program to create a GCD routine in x86 assembly language, using recursive algorithm.

Requirements and Specifications

The greatest common divisor (GCD of two positive integers m and n can be calculated recursively by the function described below in pseudocode. function GCD(m, n : integer) : integer; if n = 0 then return m; else Remainder : = m mod n;
return GCD(n, Remainder);end if;
Implement this recursive definition in assembly language. Use the stack to pass the two doubleword-size argument values. Return the value of the function in the EAX register. The procedure should remove the parameters from the stack. Test your function with a main program that inputs two integers, calls the greatest common divisor function CD, and displays the value returned.
Program Specification:
Basic I/O (using io.h)
  • Obtaining user input
  • Using an input file
  • Displaying computation results
Data Types
  • WORD
Basic Instructions and Computations
  • mov
  • cmp
  • div
  • mul
  • push
  •  рор
Write procedures
  • main procedure
  • GCD procedure
Screenshots of output
GCD routine in x86 assembly language recursive algorithm
GCD routine in x86 assembly language recursive algorithm 1
GCD routine in x86 assembly language recursive algorithm 2
Source Code
.STACK 4096
promptM BYTE 'Enter a value for m: ', 0
promptN BYTE 'Enter a value for n: ', 0
gcdTitle BYTE 'GCD result', 0
gcdMessage BYTE 'GCD(m,n) = '
gcdResult BYTE 12 DUP(0) ; place to save result string
buffer BYTE 20 DUP(?) ; place to read strings
_MainProc PROC
    input promptM, buffer, 20 ; read first number string (m)
    atod buffer ; convert string to a number
    mov M, eax ; save in variable
    input promptN, buffer, 20 ; read second number string (n)
    atod buffer ; convert string to a number
    mov N, eax ; save in variable
    push N ; pass n as second argument
    push M ; pass m as first argument
    call GCD ; calculate GCD(n, n)
    dtoa gcdResult, eax ; convert result to string
    output gcdTitle, gcdMessage ; show gcd result
    mov eax, 0 ; exit with return code 0
_MainProc ENDP
; Function GCD(m,n)
    push ebp ; save frame pointer
    mov ebp, esp ; point to top of stack
    push ebx ; save EBX
    mov eax, [ebp + 8] ; load first argument (m)
    mov ebx, [ebp + 12] ; load second argument (n)
    cmp ebx, 0 ; if n == 0
    je _return ; return m (is already in eax)
    mov edx, 0 ; clear edx to make division
    div ebx ; divide m/n
    push edx ; pass remainder as second argument
    push ebx ; pass n as first argument
    call GCD ; recurse GCD(n, remainder)
    pop ebx ; restore EBX
    pop ebp ; restore frame pointer
    ret 8 ; return and remove arguments from stack
END ; end of source code