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嘿伙计们，这是我第一次发帖，所以对我好点！我必须编写一个程序来移动和减去最终将两个数字相除。我已经做了一个可以移位加法和乘法的程序......现在只需要除法。这就是我到目前为止所拥有的。你必须制作一个带有菜单的程序，它要求用户输入 2 个输入，然后它加、减、乘然后除。

product_a_b proc near
        clc
        push ax 
        push cx
        push bx
        push si
        push dx 
        push di
        push bp

        mov si, 00h
        mov ax, 00h         
        mov bx, 00h
        mov dx, 00h
        mov cx, 00

        mov al, A[si]
        mov ah, A[si+1]
        mov dl, B[bx]
        mov dh, B[bx+1]

        mov bp, 10h

multiply:   TEST DX, 01B

        jz  skip             
        add  cx, ax

skip:       rcr  cx, 01
        rcr  dx, 01
        dec  bp               ; dec number
        jnz  multiply                           

        mov [di],dl
        mov [di+1],dh
        mov [di+2],cl
        mov [di+3],ch

        pop bp
        pop di                        
        pop dx
        pop si
        pop bx
        pop cx
        pop ax
        ret               
product_a_b endp               

QUOTIENT_A_B proc near
        clc
        push ax 
        push cx
        push bx
        push si
        push dx 
        push di
        push bp

        mov si, 00h
        mov ax, 00h         
        mov bx, 00h
        mov dx, 00h
        mov cx, 00

        mov dl, A[si]          ;ax = m 
        mov dh, A[si+1]
        mov al, B[bx]          ; dx = q
        mov ah, B[bx+1]        ; cx = a

        mov bp, 10h

divide:     rcl  cx, 01
        rcl  dx, 01  
        cmp  cx, dx         
        jb   div_add
        dec  bp               ; dec number
        jnz  divide                           

div_add:    add cx, dx
        jmp divide


        mov [di],dl
        mov [di+1],dh
        mov [di+2],cl
        mov [di+3],ch

        pop bp
        pop di                        
        pop dx
        pop si
        pop bx
        pop cx
        pop ax
        ret         
QUOTIENT_A_B endp
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1 回答 1

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像这样的东西可能会起作用：

; divide cx by dx, leaving the quotient in cx and the remainder in ax.
         mov  ax, 0
         mov  bp, 10h
divide:  sal  cx, 1
         rcl  ax, 1
         cmp  ax, dx
         jb   smaller
         sub  ax, dx
         inc  cx
smaller: dec  bp
         jnz  divide
于 2011-07-06T19:00:07.220 回答