按照Steve Johnson的指示,我没有选择“Source Mode”来跟踪程序集,事实上在push ebp之后,出现了很多(真的很多!)汇编代码,它们在选择“Source Mode”时隐藏了。所以,这种情况下是解决了,但我发现其他问题,首先请参阅我的更完整的代码:
void func();
int main(int argc, char* argv[])
{
func();
return 0;
}
void func2();
void func()
{
func2();
}
在FUNC,我声明任何局部变量,下面是当我执行FUNC()在主WinDbg的输出:
004113c3 50 push eax
0:000>
eax=0000000a ebx=7ffda000 ecx=00000000 edx=00000001 esi=01f1f760 edi=0012ff68
eip=004113c4 esp=0012fe98 ebp=0012ff68 iopl=0 nv up ei pl nz na po nc
cs=001b ss=0023 ds=0023 es=0023 fs=003b gs=0000 efl=00000202
SimpleStack!main+0x24:
004113c4 e8e0fdffff call SimpleStack!ILT+420(_func) (004111a9)
0:000> t
eax=0000000a ebx=7ffda000 ecx=00000000 edx=00000001 esi=01f1f760 edi=0012ff68
eip=004111a9 esp=0012fe94 ebp=0012ff68 iopl=0 nv up ei pl nz na po nc
cs=001b ss=0023 ds=0023 es=0023 fs=003b gs=0000 efl=00000202
SimpleStack!ILT+420(_func):
004111a9 e942020000 jmp SimpleStack!func (004113f0)
0:000> t
eax=0000000a ebx=7ffda000 ecx=00000000 edx=00000001 esi=01f1f760 edi=0012ff68
eip=004113f0 esp=0012fe94 ebp=0012ff68 iopl=0 nv up ei pl nz na po nc
cs=001b ss=0023 ds=0023 es=0023 fs=003b gs=0000 efl=00000202
SimpleStack!func:
004113f0 55 push ebp
0:000> t
eax=0000000a ebx=7ffda000 ecx=00000000 edx=00000001 esi=01f1f760 edi=0012ff68
eip=004113f1 esp=0012fe90 ebp=0012ff68 iopl=0 nv up ei pl nz na po nc
cs=001b ss=0023 ds=0023 es=0023 fs=003b gs=0000 efl=00000202
SimpleStack!func+0x1:
004113f1 8bec mov ebp,esp
0:000> t
eax=0000000a ebx=7ffda000 ecx=00000000 edx=00000001 esi=01f1f760 edi=0012ff68
eip=004113f3 esp=0012fe90 ebp=0012fe90 iopl=0 nv up ei pl nz na po nc
cs=001b ss=0023 ds=0023 es=0023 fs=003b gs=0000 efl=00000202
SimpleStack!func+0x3:
004113f3 81ecc0000000 sub esp,0C0h
0:000> t
eax=0000000a ebx=7ffda000 ecx=00000000 edx=00000001 esi=01f1f760 edi=0012ff68
eip=004113f9 esp=0012fdd0 ebp=0012fe90 iopl=0 nv up ei pl nz na po nc
cs=001b ss=0023 ds=0023 es=0023 fs=003b gs=0000 efl=00000202
SimpleStack!func+0x9:
004113f9 53 push ebx
0:000> t
eax=0000000a ebx=7ffda000 ecx=00000000 edx=00000001 esi=01f1f760 edi=0012ff68
eip=004113fa esp=0012fdcc ebp=0012fe90 iopl=0 nv up ei pl nz na po nc
cs=001b ss=0023 ds=0023 es=0023 fs=003b gs=0000 efl=00000202
SimpleStack!func+0xa:
004113fa 56 push esi
事实上,我们看到,推EBP后,它是MOV EBP,ESP,这实际上改变EBP,但有一个代码 “子ESP,0C0H” 下面的 “MOV EBP,ESP”,我知道“ sub esp,num“是为堆栈帧上的局部变量分配内存空间,但我没有在func中声明局部变量,所以我有这样的问题: ”sub esp,0C0h“是做什么的回覆? 盯着功能prolog,uf输出是:
0:000> uf func
SimpleStack!func [d:\code\simplestack\func.c @ 4]:
4 004113f0 55 push ebp
4 004113f1 8bec mov ebp,esp
4 004113f3 81ecc0000000 sub esp,0C0h
4 004113f9 53 push ebx
4 004113fa 56 push esi
4 004113fb 57 push edi
4 004113fc 8dbd40ffffff lea edi,[ebp-0C0h]
4 00411402 b930000000 mov ecx,30h
4 00411407 b8cccccccc mov eax,0CCCCCCCCh
4 0041140c f3ab rep stos dword ptr es:[edi]
5 0041140e e83dfcffff call SimpleStack!ILT+75(_func2) (00411050)
6 00411413 5f pop edi
6 00411414 5e pop esi
6 00411415 5b pop ebx
6 00411416 81c4c0000000 add esp,0C0h
6 0041141c 3bec cmp ebp,esp
6 0041141e e818fdffff call SimpleStack!ILT+310(__RTC_CheckEsp) (0041113b)
6 00411423 8be5 mov esp,ebp
6 00411425 5d pop ebp
6 00411426 c3
它没有。 “push ebp”的确如它在锡上所说的那样。请注意,eip增加1E,如果只执行了推动,它会增加1倍。显然,不止推动已经执行。 – harold 2012-03-19 11:14:27
这里还有其他事情正在发生。 EIP的差异(0041140e - 004113f0)= 1E字节。太大而不能成为单一指令。你有多个线程在运行吗? – adelphus 2012-03-19 11:14:32
@adelphus:你是对的。只需在'004113f0..0041140e'之间进行反汇编即可找到一个序列(在此使用Intel语法):'PUSH EBP; MOV EBP,ESP; PUSH; ...; SUB ESP,...';这是由编译器生成的函数入口代码(称为_function prologue_),它使代码实现了通常的C调用约定/ ABI。在函数结束/返回时,您会找到相应的块“ADD ESP,...; POP ...;离开; RET'解开了这一切。 –
2012-03-19 12:03:31