【发布时间】:2015-08-10 12:45:05
【问题描述】:
在以下使用 Ida pro 的 Hex 射线反编译的函数中:
int sub_409650()
{
int v0; // ecx@1
int result; // eax@1
bool v2; // zf@1
bool v3; // sf@1
unsigned __int8 v4; // of@1
unsigned __int16 v5; // cx@2
unsigned int v6; // ecx@2
v0 = gS1_dword_62EEA8 & 7;
result = gS1_dword_62EEA8 - v0;
v4 = __OFSUB__(gS1_dword_62EEA8 - v0, 16);
v2 = gS1_dword_62EEA8 - v0 == 16;
v3 = gS1_dword_62EEA8 - v0 - 16 < 0;
gS1_dword_62EEA8 -= v0;
gs2_dword_62EFB4 >>= v0;
if ( (unsigned __int8)(v3 ^ v4) | v2 )
{
v5 = *dword_62EFB0;
++dword_62EFB0;
v6 = (v5 << result) | gs2_dword_62EFB4;
result += 16;
gs2_dword_62EFB4 = v6;
gS1_dword_62EEA8 = result;
}
return result;
}
它调用__OFSUB__,但这有什么作用?我认为这与溢出有关 - 但如果这是真的,那么为什么不是条件:
// Checking if subtracting v0 is 16 or negative?
if ( v3 | v2 )
更新:原始 asm 是(现在重命名了一些东西):
.text:00409650 sub_409650 proc near
.text:00409650 mov eax, gBitCounter_62EEA8
.text:00409655 push esi
.text:00409656 mov esi, gFirstAudioFrameDWORD_dword_62EFB4
.text:0040965C mov ecx, eax
.text:0040965E and ecx, 7
.text:00409661 shr esi, cl
.text:00409663 sub eax, ecx
.text:00409665 cmp eax, 10h
.text:00409668 mov gBitCounter_62EEA8, eax
.text:0040966D mov gFirstAudioFrameDWORD_dword_62EFB4, esi
.text:00409673 jg short loc_4096A5
.text:00409675 mov edx, gAudioFrameDataPtr
.text:0040967B xor ecx, ecx
.text:0040967D mov cx, [edx]
.text:00409680 add edx, 2
.text:00409683 mov esi, ecx
.text:00409685 mov ecx, eax
.text:00409687 shl esi, cl
.text:00409689 mov ecx, gFirstAudioFrameDWORD_dword_62EFB4
.text:0040968F mov gAudioFrameDataPtr, edx
.text:00409695 or ecx, esi
.text:00409697 add eax, 10h
.text:0040969A mov gFirstAudioFrameDWORD_dword_62EFB4, ecx
.text:004096A0 mov gBitCounter_62EEA8, eax
.text:004096A5
.text:004096A5 loc_4096A5: ; CODE XREF: sub_409650+23j
.text:004096A5 pop esi
.text:004096A6 retn
.text:004096A6 sub_409650 endp
【问题讨论】:
-
那么为什么代码也有 v3 = gS1_dword_62EEA8 - v0 - 16
-
也想提供组装吗?
-
我在 asm 列表中进行了编辑
标签: c reverse-engineering decompiler ida