原文地址：Question about disassembler

簡介

這篇文章介紹瞭如何在不使用外掛的IDA Hex-Rays如何得到比較清晰的虛擬碼。IDA Hex-Rays功能很強大，只要你提供了足夠多的資訊，它就能產生十分簡單明瞭的程式碼。

下面我們以下面這個二進位制檔案為例：

為了方便我直接把exe檔案字尾改成jpg，下載下來把檔案字尾改回exe就行了

二進位制檔案下載地址：

步驟

開啟IDA Pro載入這個exe檔案，先按shitf + F5，新增vc32_14, vc32rtf, vc32ucrt這三個符號簽名檔案。

這個exe檔案的main函式不太好找，我們先定位到exe檔案的入口，按F5得到以下結果

signed int __usercall [email protected]<eax>(int [email protected]<ebp>, int [email protected]<esi>)
{
  char v2; // bl
  int v4; // ST14_4
  _DWORD *v5; // eax
  _DWORD *v6; // esi
  _DWORD *v7; // eax
  _DWORD *v8; // esi
  const char **v9; // edi
  int *v10; // esi
  const char **v11; // eax

  sub_4018B4();
  if ( !(unsigned __int8)__scrt_initialize_crt(1)
    || (v2 = 0, *(_BYTE *)(a1 - 25) = 0, *(_DWORD *)(a1 - 4) = 0, *(_BYTE *)(a1 - 36) = sub_401631(), dword_41CC40 == 1) )
  {
    __scrt_fastfail(7);
    goto LABEL_20;
  }
  if ( dword_41CC40 )
  {
    v2 = 1;
    *(_BYTE *)(a1 - 25) = 1;
  }
  else
  {
    dword_41CC40 = 1;
    if ( _initterm_e(&unk_415140, &unk_415158) )
    {
      *(_DWORD *)(a1 - 4) = -2;
      return 255;
    }
    _initterm(&unk_415134, &unk_41513C);
    dword_41CC40 = 2;
  }
  __scrt_release_startup_lock(*(_DWORD *)(a1 - 36));
  v5 = (_DWORD *)sub_40196C(v4);
  v6 = v5;
  if ( *v5 )
  {
    if ( (unsigned __int8)__scrt_is_nonwritable_in_current_image(v5) )
      ((void (__thiscall *)(_DWORD, _DWORD, signed int, _DWORD))*v6)(*v6, 0, 2, 0);
  }
  v7 = (_DWORD *)sub_401972();
  v8 = v7;
  if ( *v7 )
  {
    if ( (unsigned __int8)__scrt_is_nonwritable_in_current_image(v7) )
      _register_thread_local_exe_atexit_callback(*v8);
  }
  v9 = *(const char ***)sub_406ACE();
  v10 = (int *)sub_406AC8();
  v11 = (const char **)unknown_libname_31();
  a2 = main(*v10, v9, v11);
  if ( !(unsigned __int8)sub_401A94() )
LABEL_20:
    exit(a2);
  if ( !v2 )
    _cexit();
  __scrt_uninitialize_crt(1, 0);
  *(_DWORD *)(a1 - 4) = -2;
  return a2;
}
 

注意下面的程式碼

  a2 = main(*v10, v9, v11);
  if ( !(unsigned __int8)sub_401A94() )
LABEL_20:
    exit(a2);

exit函式的引數應該就是主函式的返回值。

定位到到main函式

.text:00401390                 mov     esi, eax
.text:00401392                 call    sub_406312
.text:00401392
.text:00401397                 push    eax             ; envp
.text:00401398                 push    edi             ; argv
.text:00401399                 push    dword ptr [esi] ; argc
.text:0040139B                 call    main
 

按F5，我們會得到以下程式碼

int __cdecl main(int argc, const char **argv, const char **envp)
{
  HMODULE v3; // esi
  CHAR v4; // al
  int v5; // ecx
  unsigned int v6; // esi
  char v7; // bl
  __int128 v9; // [esp+4h] [ebp-22Ch]
  int v10; // [esp+14h] [ebp-21Ch]
  int v11; // [esp+18h] [ebp-218h]
  __int16 v12; // [esp+1Ch] [ebp-214h]
  char v13; // [esp+1Eh] [ebp-212h]
  int v14; // [esp+20h] [ebp-210h]
  char v15; // [esp+24h] [ebp-20Ch]
  CHAR Buffer[256]; // [esp+124h] [ebp-10Ch]
  int v17; // [esp+224h] [ebp-Ch]
  char v18; // [esp+228h] [ebp-8h]

  v3 = GetModuleHandleA(0);
  memset(Buffer, 0, 0xFFu);
  memset(&v15, 0, 0xFFu);
  if ( !LoadStringA(v3, 0x539u, Buffer, 255) )
    return -1;
  v4 = Buffer[0];
  if ( Buffer[0] )
  {
    v5 = 0;
    do
    {
      *((_BYTE *)&v14 + ++v5 + 3) = v4 ^ 0x30;
      v4 = Buffer[v5];
    }
    while ( v4 );
  }
  memset(Buffer, 0, 0xFFu);
  if ( !LoadStringA(v3, 0x29Au, Buffer, 255) )
    return -1;
  v17 = 0;
  v18 = 0;
  v14 = 5;
  if ( RegGetValueA(-2147483647, &v15, Buffer, 0xFFFF, 0, &v17, &v14) )
    return -1;
  v6 = 0;
  v9 = xmmword_4194E0;
  v10 = 55858812;
  v7 = 114;
  v11 = 1157851502;
  v12 = 20051;
  v13 = 0;
  do
  {
    sub_401010((const char *)&unk_4194D0, v7 ^ *((_BYTE *)&v17 + v6 % (v14 - 1)));
    v7 = *((_BYTE *)&v9 + v6++ + 1);
  }
  while ( v7 );
  return 0;
}

看起來確實很亂，但我們可以幫助反彙編器給出一個稍好一點的程式碼

首先我們有兩個memsets

  memset(Buffer, 0, 0xFFu);
  memset(&v15, 0, 0xFFu);