Question

我是装配新手，并且一直坚持使用这个程序，其中使用了一个简单的添加功能，它接受两个输入并给出总和。我现在得到的输出总是0。我认为这与数据类型的处理方式有关。任何帮助表示赞赏。

注意：程序在没有功能的情况下工作正常

.386
.model flat, stdcall
option casemap:none

include \masm32\include\windows.inc
include \masm32\include\kernel32.inc
include \masm32\include\masm32.inc

includelib \masm32\lib\kernel32.lib
includelib \masm32\lib\masm32.lib



.data

msg1 db "Please enter first number ",0
msg2 db "Pleae enter second number ",0
input1 db 10 DUP(0)
input2 db 10 DUP(0)
sum dword 0
sums db 10 DUP(0)
msg3 db "The sum of your numbers is :",0
temp1 dword     0
temp2 dword     0



fsum PROTO :dword, :dword, :dword

.code


start:

    invoke StdOut , addr msg1
    invoke StdIn , addr input1,10
    invoke StdOut , addr msg2
    invoke StdIn , addr input2,10


    ;Strip CRLF
    invoke StripLF, addr input1
    invoke StripLF, addr input2


    ;string to int

    invoke atodw, addr input1
    mov temp1,eax
    invoke atodw , addr input2
    mov temp2,eax

    ;Function CALL
    invoke fsum, addr temp1,addr temp2,addr sum




    ;int to string
    invoke dwtoa,sum, addr sums

    ;Printing OUTPUT
    invoke StdOut, addr msg3
    invoke StdOut, addr sums

    invoke ExitProcess, 0

    ;Function Definition

    fsum PROC x:DWORD , y:DWORD , z:DWORD

    mov eax,x
    add eax,y
    mov z,eax

    ret
    fsum endp

end start

Answer 1

通过地址传递前两个参数毫无意义。当你只能返回结果时，输出参数也没有任何实际意义。

因此，更合理的方法是：

invoke fsum, temp1, temp2
; The sum is now in eax

...

fsum PROC x:DWORD , y:DWORD
    mov eax,x
    add eax,y
    ret 
fsum endp

如果你绝对坚持输出参数，你仍然需要做一些小改动：

; First two arguments are passed by value; last one by address
invoke fsum, temp1,temp2,addr sum

...

fsum PROC x:DWORD , y:DWORD , z:DWORD
    mov eax,x
    add eax,y
    mov edx,z       ; Get the address to write to
    mov [edx],eax   ; ..and write to it.
    ret
fsum endp

英特尔汇编程序使用功能

1 个答案: