问题描述
我目前正在做一个分配，在其中编写一个子例程，其中将2个无符号数字相乘并在DX:AX对中产生结果.但是我不能使用mul，imul，div和idiv指令.当我运行代码时，下半部分(AX寄存器)总是正确的，但DX寄存器却不正确.谁能指出我做错了正确的方向?
I'm currently working on an assignment, where I write a subroutine where 2 unsigned numbers get multiplied and yield a result in the DX:AX pair. But i cannot use the instructions mul, imul, div, and idiv. When i run my code, the bottom half (the AX register) is always correct, but the DX register is not. Can anyone point me in the right direction as to what I am doing wrong?
;-----------------------------------------------------------
;
; Program: MULTIPLY
;
; Function: Multiplies two 16 bit unsigned values ...
; .... duplicating the MUL instruction
;
; Input: The two values to be multiplied are passed on the stack
; The code conforms to the C/C++ calling sequence
;
; Output: The 32 bit result is returned in the dx:ax pair
; Registers required by C/C++ need to be saved and restored
;
; Owner: Andrew F.
;
; Changes: Date Reason
; ------------------
; 07/20/2013 Original version
;
;
;---------------------------------------
.model small
.8086
public _multiply
.data
;---------------------------------------
; Multiply data
;---------------------------------------
.code
;---------------------------------------
; Multiply code
;---------------------------------------
_multiply:
push bp ; save bp
mov bp,sp ; anchor bp into the stack
mov ax,[bp+4] ; load multiplicand from the stack
mov dx,[bp+6] ; load multiplier from the stack
push bx
push cx
push di
;---------------------------------------
; copy ax to cx, and dx to bx
;---------------------------------------
mov cx,ax ;using bx and cx as my inputs
mov bx,dx
;---------------------------------------
; Check for zeros, zero out ax and dx
;---------------------------------------
start:
xor ax,ax ; check for multiplication by zero
mov dx,ax ; and zero out ax and dx
mov di,cx ;
or di,bx ;
jz done ;
mov di,ax ; DI used for reg,reg adc
;---------------------------------------
; loop / multiply algorithm
;---------------------------------------
loopp:
shr cx,1 ; divide by two, bottom bit moved to carry flag
jnc skipAddToResult ;no carry -> just add to result
add ax,bx ;add bx to ax
adc dx,di ;add the carry to dx
skipAddToResult:
add bx,bx ;double bx current value
or cx,cx ; zero check
jnz loopp ; if cx isnt zero, loop again
;---------------------------------------
; Restore register values, return
;---------------------------------------
done:
pop di ;restore di
pop cx ;restore cx
pop bx ;restore bx
pop bp ; restore bp
ret ; return with result in dx:ax
;
end ; end source code
;---------------------------------------
推荐答案
添加另一个移位的值bx时，您奇怪地使用了di.您的算法似乎是这样的:
You are strangely using di when adding another shifted value of bx. Your algorithm seems to be like this:
- 收集值，并将其放入BX和CX中.
- 而CX> 0:
- Gather values, put them into BX and CX.
- While CX>0:
- 向右移动CX.
- 如果移位后的比特为1，则将BX添加到AX，然后用进位将DI(DI中为零?)添加到DX.
- 将BX添加到BX.
在CX每次右移之后，您将丢失DI:BX的左移.当DI保持为零时，您仅移动BX(我将使用shl bx,1而不是add bx,bx)，因此，当BX超过16位时，您将丢失应该转到DX的位.要解决此问题，请对DI使用旋转进位.
You are missing a left-shift of DI:BX after each right shift of CX. You are shifting only BX (and I'd use shl bx,1 instead of add bx,bx) while DI remains at zero, therefore when BX exceeds 16 bits, you lose bits that should go to DX. To remedy, use rotate through carry against DI.
loopp:
shr cx,1 ; divide by two, bottom bit moved to carry flag
jnc skipAddToResult ;no carry -> just add to result
add ax,bx ;add bx to ax
adc dx,di ;add the carry to dx
skipAddToResult:
shl bx,1 ;double bx current value
rcl di,1 ; and put overflow bits to DI
; this together doubles the number in DI:BX
or cx,cx ; zero check
jnz loopp ; if cx isnt zero, loop again
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