[Assembly]GCD

GCD，最大公因數（Greatest Common Divisor，簡寫為G.C.D.；或Highest Common Factor，簡寫為H.C.F.），指某幾個整數共有因數中最大的一個。

想法: 從C/C++程式碼轉換成組合語言。

迴圈版本

INCLUDE Irvine32.inc
.data
myMessage BYTE "Please enter two integer number.",0dh,0ah,0
intVal1 SDWORD ?
intVal2 SDWORD ?
.code
main PROC
    call Clrscr
    mov  edx,OFFSET myMessage
    call WriteString
 
 
    call ReadInt
    mov intVal1,eax
    cdq                 ;abs
    xor eax , edx       ;abs
    sub eax , edx       ;abs
    mov ebx , eax
 
    call ReadInt
    mov intVal1,eax
    cdq                 ;abs
    xor eax , edx       ;abs
    sub eax , edx       ;abs
    mov ecx , eax
 
    call GCD
    call WriteInt
    exit
main ENDP
;---------------------------------------------------------
GCD PROC 
;   Yo this func can help U to find the GCD from two Integer
;   Receives: intVal1 in ebx , intVal2 in ecx
;   Returns: GCD in eax
;---------------------------------------------------------
    mov eax , ebx
L1: cdq
    idiv ecx
    mov eax , ecx 
    mov ecx , edx 
    cmp ecx , 0
    jg L1
    ret
GCD ENDP
END main

遞迴版本

INCLUDE Irvine32.inc
.data
myMessage BYTE "Please enter two integer number.",0dh,0ah,0
intVal1 SDWORD ?
intVal2 SDWORD ?
.code
main PROC
    call Clrscr
    mov ecx , 5
L1:
    mov  edx,OFFSET myMessage
    call WriteString
     
 
 
    call ReadInt
    mov intVal1,eax
    cdq                 ;abs
    xor eax , edx       ;abs
    sub eax , edx       ;abs
    push eax
 
    call ReadInt
    mov intVal1,eax
    cdq                 ;abs
    xor eax , edx       ;abs
    sub eax , edx       ;abs
    push eax
 
    call GCD
    call WriteInt
    call CrLf
    Loop L1
 
    exit
main ENDP
;---------------------------------------------------------
GCD PROC 
;   Yo this func can help U to find the GCD from two Integer Implentnt by Recursion
;
;   int gcd(int m, int n) { 
;        if(n == 0) 
;             return m; 
;        else
;             return gcd(n, m % n); 
;   }
;
;   Receives: intVal1 in ebx , intVal2 in ecx
;   Returns: GCD in eax
;---------------------------------------------------------
    push ebp
    mov ebp , esp
 
    mov eax , [ebp+8]       ;n
    cmp eax,0               ; if n == 0
    ja L1
    mov eax , [ebp+12]      ;m
    je Quit
 
L1:
    mov eax , [ebp+12]
    cdq
    mov ebx , [ebp+8]
    idiv ebx                
    push [ebp+8]            ; n
    push edx                ; m % n
    call GCD
 
Return:
 
Quit:
    pop ebp
    ret 8
GCD ENDP
END main