Question

我正在试图弄清楚如何对程序中的角色进行排序。有些原因我认为我的排序搞乱了，因为当我把它放在程序中时就崩溃了，但是当它不在程序中它运行得很好。

现在这里是带有排序参数的cpp文件：

    extern "C" void Sort (char [] [11], char [], double [], long);

这是我在汇编中的排序

_Sort proc
    mov eax, [esp + 8]
    cmp eax, 1
    jle L1
    push ebp

    mov ebp, esp
    mov edx, 1
    imul edx, Row
    mov Count, edx

    imul edx, Column
    push edx 
    cld

L1:
    call _comp
    cmp ebx, 1
    jz L1
    add esp, 4
    pop ebp
L2:
    ret
_Sort endp

_Comp Proc
    mov edx, 0
    mov Count, 0;

L3:
    mov edi, offset _pArray1
    mov esi, pArray
    add esi, edx
    mov ecx, Column
    rep movsb

    add edx, Column
    inc Count
    mov edi, offset _pArray2
    mov ecx, Column
    rep movsb

    mov edi, offset _pArray1
    mov esi, offset _pArray2
    mov ecx, Column
    repz cmpsb

    jge L4
    call _Switch
    mov ebx, 1
    jmp L5
L4:
    mov ebx, 0
    mov eax, Done
    sub eax, Column
    cmp edx, eax
    jz L5
    jmp L3

L5:
    ret

_Comp endp

_Switch proc

    mov edi, pArray
    add edi, edx
    sub edi, Column
    mov esi, offset _pArray2

    mov ecx, Column
    rep movsb
    mov esi, offset _pArray1
    mov ecx, Column
    rep movsb

    mov esi, [ebp + 12]
    add esi, Count
    dec esi

    mov cl, [esi]
    xchg cl, [esi +1]
    mov [esi], cl

    ret

_Switch endp

Answer 1

我只是指出了一些在阅读代码时显而易见的问题，所以我看起来并不多，特别是当你使用既不显示也不看似初始化的变量时。

首先调用函数时的堆栈：

 sp+ 0 = caller return address
 sp+ 4 = char [] [11]
 sp+ 8 = char []
 sp+ c = double []
 sp+10 = long

所以你开始你的代码，你检查的第一件事是指针char []等于1.我想这不是你想要的。

mov eax, [esp + 8]
cmp eax, 1

如果这是真的，可能不会发生，但它可能为null，那么你执行这段代码：

L1:
call _comp   ; we call _comp which uses ebp from the parent function, so it is to be considered not as initialzed *BAM*
cmp ebx, 1   ; why should ebx be 1? You never initilized it, but maybe it is a random value frmo the caller?
jz L1
add esp, 4   ; you are now corrupting the stack
pop ebp      ; corrupting it even more

L2:
ret          ; and jump back to  where char[] points to. *BAM*

现在假设没有采取分支，我们继续在这里：

    push ebp
    mov ebp, esp     ; setup the stackframe, which is ok.

    mov edx, 1
    imul edx, Row    ; Where does Row com from? Is it initlized? Where?
    mov Count, edx   ; Save the value in Count, but...

    imul edx, Column
    push edx         ; ?
    cld

L1:
    call _comp      ; ... here we call _comp which first instruction is, to initialize Count again.

....

所以我建议重新考虑这段代码，添加注释以便知道它应该做什么，然后在调试器中运行它来检查它是否正在按照你认为的那样做。

对程序集中的字符串进行排序

1 个答案: