零之启之反破解测试

筱玉破解吧

几分钟弄好的，没什么难度，就加密了一些代码，练手很适合，来挑战试试
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linjicheng88

// positive sp value has been detected, the output may be wrong!
unsigned __int32 __usercall sub_D00656@<eax>(int _EDX@<edx>, char a2@<bh>, int a3, int a4, int a5, int a6, int a7, int a8, int a9, int a10, int a11, int a12, int a13, int a14, int a15, int a16, int a17, int a18, int a19, int a20, int a21, int a22, int a23, int a24, int a25, int a26, int a27, int a28, int a29, int a30, int a31, int a32, int a33, int a34, int a35, int a36, int a37)
{
  unsigned int v37; // ebx
  unsigned int v39; // kr00_4

  v39 = __readeflags();
  _EAX = (unsigned __int16)_EDX;
  _ECX = _byteswap_ulong((unsigned __int8)~a2 + 2) + 4;
  LOWORD(_ECX) = _ECX + 1;
  LOBYTE(_ECX) = _byteswap_ulong(_ECX);
  __asm
  {
    arpl    bx, ax
    arpl    ax, ax
    rcr     dl, cl
  }
  LOBYTE(_ECX) = _ECX + 2;
  __asm { rcr     edx, cl }
  v37 = __ROR4__(a37, 3) + 1;
  BYTE1(v37) = 0;
  return _byteswap_ulong(v37) & ((_EAX | (1 << __ROL4__(_EDX, 37))) + 1);

易语言灬魅影

void __usercall sub_C78EC2(int a1@<eax>, int a2@<edx>, int a3@<ecx>, int _EBX@<ebx>, char a5@<efl>, int a6)
{
  unsigned __int16 v6; // di
  unsigned int v9; // kr00_4
  int v12; // ebx
  unsigned __int32 v13; // ebx
  _DWORD v14[57]; // [esp-DCh] [ebp-E4h] BYREF

  v6 = a3;
  v14[54] = a1;
  BYTE1(a1) = a5;
  v14[52] = a3;
  _CX = a1;
  v14[51] = a2;
  _AX = _byteswap_ulong(v6);
  v9 = __readeflags();
  v14[49] = v9;
  LOWORD(_EBX) = __ROR2__(_EBX, _CX);
  __asm { rcr     ax, 63h }
  v14[48] = 0;
  __asm { rcr     bx, cl }
  _EBX = _EBX & ~(1 << a6);
  __asm { rcr     ebx, cl }
  _BitScanForward((unsigned __int16 *)&_CX, -(_CX | v6));
  v12 = _CX;
  _bittest(&v12, (unsigned int)v14);
  v13 = _byteswap_ulong(__ROR4__(1 - __ROL4__(a6, 1) - 754939151, 1) - 159392616) + 1;
  __asm { retn }
}

易语言灬魅影

// positive sp value has been detected, the output may be wrong!
unsigned __int32 __usercall sub_D00656@<eax>(int _EDX@<edx>, char a2@<bh>, int a3, int a4, int a5, int a6, int a7, int a8, int a9, int a10, int a11, int a12, int a13, int a14, int a15, int a16, int a17, int a18, int a19, int a20, int a21, int a22, int a23, int a24, int a25, int a26, int a27, int a28, int a29, int a30, int a31, int a32, int a33, int a34, int a35, int a36, int a37)
{
  unsigned int v37; // ebx
  unsigned int v39; // kr00_4

  v39 = __readeflags();
  _EAX = (unsigned __int16)_EDX;
  _ECX = _byteswap_ulong((unsigned __int8)~a2 + 2) + 4;
  LOWORD(_ECX) = _ECX + 1;
  LOBYTE(_ECX) = _byteswap_ulong(_ECX);
  __asm
  {
    arpl    bx, ax
    arpl    ax, ax
    rcr     dl, cl
  }
  LOBYTE(_ECX) = _ECX + 2;
  __asm { rcr     edx, cl }
  v37 = __ROR4__(a37, 3) + 1;
  BYTE1(v37) = 0;
  return _byteswap_ulong(v37) & ((_EAX | (1 << __ROL4__(_EDX, 37))) + 1);
}