将汇编代码反向工程为c代码

I think this is actually a pretty simple problem. I have to reverse engineer this assembly code to c code. I'll also provide what I think is going on so you can hopefully point to where I went wrong and I can learn from my mistakes now.

我认为这实际上是一个非常简单的问题。我必须将此汇编代码反向工程为c代码。我还会提供我认为正在发生的事情,这样你就可以指出我哪里出错了,现在我可以从错误中吸取教训。

.LFBO
    pushq   %rbp
    movq    %rsp,%rbp
    movl    %edi,-4(%rbp)
    movl    %esi,-8(%rbp)
    movl    -4(%rbp),%eax
    compl   -8(%rbp),%eax
    jg      .L2
    movl    -8(%rbp),%eax
    jmp     .L3
.L2:
    movl    -4(%rbp),%eax
.L3:
    popq    %rbp
    ret

So this is what I think is going on with this: the first two lines after .LFBO:

所以这就是我认为正在发生的事情:.LFBO之后的前两行:

pushq   %rbp
movq    %rsp,%rbp

are just setting up the stack for the execution that is about to follow.

只是为即将执行的执行设置堆栈。

movl    %edi,-4(%rbp)

is grabbing the first variable, call it x

抓住第一个变量,称之为x

movl    %esi,-8(%rbp)

is grabbing the second variable call it y

抓住第二个变量称之为y

movl    -4(%rbp),%eax

is grabbing x to be compared in the next line

抓住x在下一行进行比较

compl   -8(%rbp),%eax

compares the variables x and y by computing x-y

通过计算x-y来比较变量x和y

jg      .L2

says jump to .L2 if x > y

如果x> y,跳转到.L2

if x <= y then compute the next lines without jumping to .L2

如果x <= y则计算下一行而不跳转到.L2

movl    -8(%rbp),%eax

copy x = y

复制x = y

jmp     .L3

jump to .L3

跳到.L3

if x > y at the jg line then you jump to .L2: and complete this line

如果x> y在jg行,那么你跳转到.L2:并完成这一行

movl    -4(%rbp),%eax

this is where I realized I was really confused. It looks to me that you're copying x to x then .L3 is completed and I think x is returned

这是我意识到我真的很困惑的地方。它在我看来你正在将x复制到x然后.L3已经完成,我认为x被返回

3 个解决方案

#1

Don't overthink it. Just gradually replace the assembly with C. Here is a possible sequence of transformations.

不要过度思考它。只需逐渐用C替换组件。这是一个可能的转换序列。

.LFBO
    pushq   %rbp
    movq    %rsp,%rbp
    movl    %edi,-4(%rbp)
    movl    %esi,-8(%rbp)
    movl    -4(%rbp),%eax
    compl   -8(%rbp),%eax
    jg      .L2
    movl    -8(%rbp),%eax
    jmp     .L3
.L2:
    movl    -4(%rbp),%eax
.L3:
    popq    %rbp
    ret

----

int LFBO (int edi, int esi)
{
    rbp = rsp
    [rbp - 4] = edi
    [rbp - 8] = esi
    eax = [rbp - 4]
    if (eax > [rbp - 8]) goto L2
    eax = [rbp - 8]
    goto L3
L2:
    eax = [rbp - 4]
L3:
    return eax
}

----

int LFBO (int edi, int esi)
{
    int eax;

    eax = edi;
    if (eax > esi) goto L2;
    eax = esi;
    goto L3;
L2:
    eax = edi;
L3:
    return eax;
}

----    

int LFBO (int edi, int esi)
{
    int eax;

    eax = edi;
    if (eax <= esi) {
        eax = esi;
    }
    else {
        eax = edi;
    }
    return eax;
}

----

int LFBO (int edi, int esi)
{
    if (edi <= esi) {
        return esi;
    }
    else {
        return edi;
    }
}

----

int LFBO (int x, int y)
{
    if (x <= y) {
        return y;
    }
    else {
        return x;
    }
}

----

int LFBO (int x, int y)
{
    return (x > y) ? x : y;
}

You can apply this strategy to any piece of assembly. Here I took the time to detail the various transformations. With practice you can get to the end result a lot quicker.

您可以将此策略应用于任何装配体。在这里,我花时间详细介绍了各种转换。通过练习,您可以更快地获得最终结果。

#2

LFB0(int x, int y){
    if (x<=y){
        x = y;
    }else{
        x = x;
    }
    return(x);
}

This is what I think we determined to be correct, with the help of the guys in the comments.

这是我认为我们在评论中的帮助下确定的正确性。

#3

.LFBO
    pushq   %rbp                prolog
    movq    %rsp,%rbp           prolog
    movl    %edi,-4(%rbp)       [ebp-4] = edi
    movl    %esi,-8(%rbp)       [ebp-8] = esi
    movl    -4(%rbp),%eax       eax = [ebp-4] ie edi
    compl   -8(%rbp),%eax       cmp eax with [ebp-8] ie esi 
    jg      .L2                 ;jg requires <= 
    movl    -8(%rbp),%eax       so cutting the junk 
    jmp     .L3                 this effectively becomes
.L2:
    movl    -4(%rbp),%eax       ( edi <= esi ) ? { eax = esi } : { eax= edi} ; return;
.L3:
    popq    %rbp                epilog
    ret                         epilog

testing the hypothesis

检验假设

lets compile the code in vc and test should compile unoptimized else 
clever compiler will cast away everything do 
/O1 push 10 pop eax retn;
/O2 mov eax ,10  ret

int main(void) {
int edi=8,esi=10;
if ( edi <= esi) { return esi; } else { return edi;}
}

disassembling the result

拆解结果

0:000> uf @eip
image00400000+0x1000:
00401000 55              push    ebp
00401001 8bec            mov     ebp,esp
00401003 83ec08          sub     esp,8
00401006 c745fc08000000  mov     dword ptr [ebp-4],8
0040100d c745f80a000000  mov     dword ptr [ebp-8],0Ah
00401014 8b45fc          mov     eax,dword ptr [ebp-4]
00401017 3b45f8          cmp     eax,dword ptr [ebp-8]
0040101a 7f07            jg      image00400000+0x1023 (00401023)

image00400000+0x101c:
0040101c 8b45f8          mov     eax,dword ptr [ebp-8]
0040101f eb05            jmp     image00400000+0x1026 (00401026)

image00400000+0x1023:
00401023 8b45fc          mov     eax,dword ptr [ebp-4]

image00400000+0x1026:
00401026 8be5            mov     esp,ebp
00401028 5d              pop     ebp
00401029 c3              ret
0:000>

#1