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我的实施结果有问题。如果你追踪它,它似乎是正确的。但是,它给出了错误的结果。如果你输入 99 和 99,它会给出 5318,但它应该给出 9801。

顺便说一句,程序接受两个 2 位数字并通过重复添加被乘数(第一个输入)来相乘,直到它满足乘数中的值(第二个输入)

我已经尝试定位问题,似乎当我使用字长除法时,它会给出随机值。word-size 最大为 65,656,但超过 255 时结果会损坏。为什么?

请注意第一个选项,即重复加法运算的乘法。

非常感谢您的帮助。

   section .data
msg db "Menu: "
msgLen equ $ -msg
msg2 db "[1]Multiplication by repeated addition"
msgLen2 equ $ -msg2
msg3 db "[2]Division by repeated subtraction"
msgLen3 equ $ -msg3
msg4 db "[3]Exit"
msgLen4 equ $ -msg4
msg5 db "Enter two numbers: "
msgLen5 equ $ -msg5
line db "", 10
result dw 0
ten dw 10
quo1 dw 0
quo2 dw 0
quo3 dw 0
rem1 dw 0
rem2 dw 0
rem3 dw 0
temp1 db 0
temp2 db 0
temp3 db 0

   section .bss
choice resb 1
num1a resw 1
num1b resw 1
num2a resw 1
num2b resw 1


   section .text
global _start

  _start:
  do_while:
mov eax, 4
mov ebx, 1
mov ecx, msg
mov edx, msgLen
int 80h

mov eax, 4
mov ebx, 1
mov ecx, line
mov edx, 1
int 80h

mov eax, 4
mov ebx, 1
mov ecx, msg2
mov edx, msgLen2
int 80h

mov eax, 4
mov ebx, 1
mov ecx, line
mov edx, 1
int 80h

mov eax, 4
mov ebx, 1
mov ecx, msg3
mov edx, msgLen3
int 80h

mov eax, 4
mov ebx, 1
mov ecx, line
mov edx, 1
int 80h

mov eax, 4
mov ebx, 1
mov ecx, msg4
mov edx, msgLen4
int 80h

mov eax, 4
mov ebx, 1
mov ecx, line
mov edx, 1
int 80h

mov eax, 3
mov ebx, 0
mov ecx, choice
mov edx, 2
int 80h

sub byte [choice], 30h

cmp byte [choice], 1
je menu1
;cmp byte [choice], 2
;je menu2
cmp byte [choice], 3
je exit
jg do_while
jl do_while

   menu1:
mov eax, 4
mov ebx, 1
mov ecx, msg5
mov edx, msgLen5
int 80h

mov eax, 3
mov ebx, 0
mov ecx, num1a
mov edx, 1
int 80h

mov eax, 3
mov ebx, 0
mov ecx, num1b
mov edx, 2
int 80h

mov eax, 3
mov ebx, 0
mov ecx, num2a
mov edx, 1
int 80h

mov eax, 3
mov ebx, 0
mov ecx, num2b
mov edx, 2
int 80h

sub word [num1a], 30h       ;conversion
sub word [num1b], 30h
sub word [num2a], 30h
sub word [num2b], 30h

push result
push word [num1a]
push word [num1b]
push word [num2a]
push word [num2b]
call func1

mov ax, [result]                ;9801
mov dx, 0
mov bx, 10
div bx

mov word [quo1], ax             ;980
mov word [rem1], dx             ;1

mov ax, [quo1]              
mov dx, 0
mov bx, 10
div bx

mov word [quo2], ax             ;98
mov word [rem2], dx             ;0

mov ax, [quo2]
mov dx, 0
mov bx, 10
div bx

mov word [quo3], ax             ;9
mov word [rem3], dx             ;8


add word [quo3], 30h
add word [rem3], 30h
add word [rem2], 30h
add word [rem1], 30h

mov eax, 4
mov ebx, 1
mov ecx, quo3
mov edx, 1
int 80h

mov eax, 4
mov ebx, 1
mov ecx, rem3
mov edx, 1
int 80h

mov eax, 4
mov ebx, 1
mov ecx, rem2
mov edx, 1
int 80h

mov eax, 4
mov ebx, 1
mov ecx, rem1
mov edx, 1
int 80h



mov eax, 4
mov ebx, 1
mov ecx, line
mov edx, 1
int 80h

jmp do_while

   func1:
mov ebp, esp

mov ax, [ebp+10]        ;90
mov dx, 0
mul word [ten]
mov [ebp+10], ax

mov ax, [ebp+8]         ;9
add word [ebp+10], ax           ;99 on [ebp+10]

mov ax, [ebp+6]         ;90
mov dx, 0
mul word [ten]
mov [ebp+6], ax

mov ax, [ebp+4]         ;9
add word [ebp+6], ax            ;99 on [ebp+6]

   while1:
mov ax, [ebp+10]
add ax, [ebp+10]

dec word [ebp+6]
cmp word [ebp+6], 1
jne while1

   end1:                                    
mov ebx, [ebp+12]
mov [ebx], ax
ret 12

   exit:
mov eax, 1
mov ebx, 0
int 80h
4

1 回答 1

0

ax下面显示的循环不正确,因为它会在每次迭代开始时重置 的值。因此,在退出循环时,ax将包含[ebp+10]乘以 2 的原始值,而不管迭代次数如何。

while1:
  mov ax, [ebp+10]
  add ax, [ebp+10]

  dec word [ebp+6]
  cmp word [ebp+6], 1
  jne while1

好的,让我们看看你的结果 5318。据说,这是[ebp+10];值的两倍。所以[ebp+10]会有值 2659,即十六进制的 0x0A63。这向我表明,高字节num1bnum2b包含换行符(ASCII 代码 0x0A)。

所以你需要修复你的循环。像这样的东西应该可以工作:

mov ax, 0
while1:
  add ax, [ebp+10]

而且你还需要清除num1a..num2b.

于 2013-08-12T08:05:02.107 回答