汇编如何在没有 ZF 标志访问的情况下将 JNE 转换为 C 代码
ASM 到 C 代码的模拟即将完成..只是试图解决这些第二遍问题。
假设我得到了这个 ASM 函数,
401040 MOV EAX,DWORD PTR [ESP+8]
401044 MOV EDX,DWORD PTR [ESP+4]
401048 PUSH ESI
401049 MOV ESI,ECX
40104B MOV ECX,EAX
40104D DEC EAX
40104E TEST ECX,ECX
401050 JE 401083
401052 PUSH EBX
401053 PUSH EDI
401054 LEA EDI,[EAX+1]
401057 MOV AX,WORD PTR [ESI]
40105A XOR EBX,EBX
40105C MOV BL,BYTE PTR [EDX]
40105E MOV ECX,EAX
401060 AND ECX,FFFF
401066 SHR ECX,8
401069 XOR ECX,EBX
40106B XOR EBX,EBX
40106D MOV BH,AL
40106F MOV AX,WORD PTR [ECX*2+45F81C]
401077 XOR AX,BX
40107A INC EDX
40107B DEC EDI
40107C MOV WORD PTR [ESI],AX
40107F JNE 401057
401081 POP EDI
401082 POP EBX
401083 POP ESI
401084 RET 8
我的程序将为其创建以下内容。
int Func_401040() {
regs.d.eax = *(unsigned int *)(regs.d.esp+0x00000008);
regs.d.edx = *(unsigned int *)(regs.d.esp+0x00000004);
regs.d.esp -= 4;
*(unsigned int *)(regs.d.esp) = regs.d.esi;
regs.d.esi = regs.d.ecx;
regs.d.ecx = regs.d.eax;
regs.d.eax--;
if(regs.d.ecx == 0)
goto label_401083;
regs.d.esp -= 4;
*(unsigned int *)(regs.d.esp) = regs.d.ebx;
regs.d.esp -= 4;
*(unsigned int *)(regs.d.esp) = regs.d.edi;
regs.d.edi = (regs.d.eax+0x00000001);
regs.x.ax = *(unsigned short *)(regs.d.esi);
regs.d.ebx ^= regs.d.ebx;
regs.h.bl = *(unsigned char *)(regs.d.edx);
regs.d.ecx = regs.d.eax;
regs.d.ecx &= 0x0000FFFF;
regs.d.ecx >>= 0x00000008;
regs.d.ecx ^= regs.d.ebx;
regs.d.ebx ^= regs.d.ebx;
regs.h.bh = regs.h.al;
regs.x.ax = *(unsigned short *)(regs.d.ecx*0x00000002+0x0045F81C);
regs.x.ax ^= regs.x.bx;
regs.d.edx++;
regs.d.edi--;
*(unsigned short *)(regs.d.esi) = regs.x.ax;
JNE 401057
regs.d.edi = *(unsigned int *)(regs.d.esp);
regs.d.esp += 4;
regs.d.ebx = *(unsigned int *)(regs.d.esp);
regs.d.esp += 4;
label_401083:
regs.d.esi = *(unsigned int *)(regs.d.esp);
regs.d.esp += 4;
return 0x8;
}
由于 JNE 401057
不使用 CMP
或 TEST
我如何修复在 C 代码中使用它?
ASM to C Code emulating nearly done.. just trying to solve these second pass problems.
Lets say I got this ASM function
401040 MOV EAX,DWORD PTR [ESP+8]
401044 MOV EDX,DWORD PTR [ESP+4]
401048 PUSH ESI
401049 MOV ESI,ECX
40104B MOV ECX,EAX
40104D DEC EAX
40104E TEST ECX,ECX
401050 JE 401083
401052 PUSH EBX
401053 PUSH EDI
401054 LEA EDI,[EAX+1]
401057 MOV AX,WORD PTR [ESI]
40105A XOR EBX,EBX
40105C MOV BL,BYTE PTR [EDX]
40105E MOV ECX,EAX
401060 AND ECX,FFFF
401066 SHR ECX,8
401069 XOR ECX,EBX
40106B XOR EBX,EBX
40106D MOV BH,AL
40106F MOV AX,WORD PTR [ECX*2+45F81C]
401077 XOR AX,BX
40107A INC EDX
40107B DEC EDI
40107C MOV WORD PTR [ESI],AX
40107F JNE 401057
401081 POP EDI
401082 POP EBX
401083 POP ESI
401084 RET 8
My program would create the following for it.
int Func_401040() {
regs.d.eax = *(unsigned int *)(regs.d.esp+0x00000008);
regs.d.edx = *(unsigned int *)(regs.d.esp+0x00000004);
regs.d.esp -= 4;
*(unsigned int *)(regs.d.esp) = regs.d.esi;
regs.d.esi = regs.d.ecx;
regs.d.ecx = regs.d.eax;
regs.d.eax--;
if(regs.d.ecx == 0)
goto label_401083;
regs.d.esp -= 4;
*(unsigned int *)(regs.d.esp) = regs.d.ebx;
regs.d.esp -= 4;
*(unsigned int *)(regs.d.esp) = regs.d.edi;
regs.d.edi = (regs.d.eax+0x00000001);
regs.x.ax = *(unsigned short *)(regs.d.esi);
regs.d.ebx ^= regs.d.ebx;
regs.h.bl = *(unsigned char *)(regs.d.edx);
regs.d.ecx = regs.d.eax;
regs.d.ecx &= 0x0000FFFF;
regs.d.ecx >>= 0x00000008;
regs.d.ecx ^= regs.d.ebx;
regs.d.ebx ^= regs.d.ebx;
regs.h.bh = regs.h.al;
regs.x.ax = *(unsigned short *)(regs.d.ecx*0x00000002+0x0045F81C);
regs.x.ax ^= regs.x.bx;
regs.d.edx++;
regs.d.edi--;
*(unsigned short *)(regs.d.esi) = regs.x.ax;
JNE 401057
regs.d.edi = *(unsigned int *)(regs.d.esp);
regs.d.esp += 4;
regs.d.ebx = *(unsigned int *)(regs.d.esp);
regs.d.esp += 4;
label_401083:
regs.d.esi = *(unsigned int *)(regs.d.esp);
regs.d.esp += 4;
return 0x8;
}
Since JNE 401057
doesn't use the CMP
or TEST
How do I fix that use this in C code?
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最近修改标志的指令是 dec,它在操作数达到 0 时设置 ZF。因此,jne 大约相当于 if (regs.d. edi!= 0)转到label_401057;。
顺便说一句:
ret 8
不等于return 8
。ret
指令的操作数是返回时添加到 ESP 的字节数。 (它通常用于清理堆栈。)这有点像,只是半明显地,这在 C 中不起作用——
return
使得它后面的任何代码都无法访问。这实际上是调用约定的一部分 -
[ESP+4]
和[ESP+8]
是传递给函数的参数,而ret< /code> 正在清理它们。这不是通常的 C 调用约定;考虑到该函数需要 ECX 中的值,它看起来更像是 fastcall 或 thiscall。
The most recent instruction that modified flags is the
dec
, which sets ZF when its operand hits 0. So thejne
is about equivalent toif (regs.d.edi != 0) goto label_401057;
.BTW:
ret 8
isn't equivalent toreturn 8
. Theret
instruction's operand is the number of bytes to add to ESP when returning. (It's commonly used to clean up the stack.) It'd be kinda likeexcept that semi-obviously, this won't work in C -- the
return
makes any code after it unreachable.This is actually a part of the calling convention --
[ESP+4]
and[ESP+8]
are arguments passed to the function, and theret
is cleaning those up. This isn't the usual C calling convention; it looks more like fastcall or thiscall, considering the function expects a value in ECX.