我在以下功能的LOBYTE / HIBYTE / LODWORD行上收到lvalue required as left operand error
:
typedef DWORD _DWORD;
typedef BYTE _BYTE;
typedef WORD _WORD;
signed int __stdcall checkSerial(int serial, int serialLength)
{
int v2; // ecx@1
int v3; // ecx@3
int v4; // ecx@5
int v5; // ecx@7
int v6; // dx@9
signed int v7; // ecx@10
char v8; // dl@11
__int16 v9; // cx@13
signed int result; // eax@21
v2 = 0;
do
{
*(_DWORD *)(v2 + serial) ^= 0x1234567u;
*(_BYTE *)(v2 + serial) &= 0xEu;
v2 += 4;
}
while ( v2 != 8 );
v3 = 0;
do
*(_BYTE *)(serial + 8) += *(_BYTE *)(v3++ + serial);
while ( v3 != serialLength );
v4 = 0;
do
{
*(_DWORD *)(v4 + serial) ^= 0x89ABCDEu;
*(_BYTE *)(v4 + serial) &= 0xEu;
v4 += 4;
}
while ( v4 != 8 );
v5 = 0;
do
*(_BYTE *)(serial + 9) += *(_BYTE *)(v5++ + serial);
while ( v5 != serialLength );
LOBYTE(v6) = *(_BYTE *)(serial + 9);
HIBYTE(v6) = *(_BYTE *)(serial + 8);
if ( v6 != 0x42DE )
goto LABEL_25;
v7 = 9;
do
{
v8 = *(_BYTE *)(v7 + serial) ^ *(_BYTE *)(v7 + serial);
--v7;
}
while ( (_WORD)v7 );
if ( (*(_WORD *)(serial + 8) ^ 0xEEEE) != 0x30AC )
goto LABEL_25;
LOBYTE(v9) = *(_BYTE *)serial;
HIBYTE(v9) = *(_BYTE *)(serial + 1);
if ( v9 != 0xB008u )
goto LABEL_25;
if ( *(_DWORD *)serial != 0x7A81B008 )
goto LABEL_25;
LOWORD(v7) = 0;
do
{
v8 ^= 0xAu;
v7 += 4;
}
while ( v7 <= 12 );
if ( *(_DWORD *)(serial + 4) != 0x388DBF02
|| *(_BYTE *)(serial + 5) != 191
|| *(_BYTE *)(serial + 6) != 141
|| *(_BYTE *)(serial + 5) != 191 )
LABEL_25:
result = 1;
else
result = 0;
return result;
}
LOBYTE(v9)= *(_ BYTE *)串行;
HIBYTE(v9)= *(_ BYTE *)(序列+ 1);
例如这行引起错误-但是为什么呢? LOBYTE被定义为LOBYTE(x)((BYTE)(x)),所以为什么我不能只转换它?
最佳答案
修改值子集的方法是使用布尔运算符,例如:
v9 &= ~255; // mask off low byte (set to 0)
v9 |= *(_BYTE*)serial; // or in the new low byte value
但是,由于您要修改一个16位值的两个字节,所以最简单的方法是只在一个操作中分配它:
v9 = (*(_BYTE*)(*serial + 1) << 8) | (*(_BYTE*)serial);
关于c++ - 需要左值作为赋值左操作数-LOBYTE/HIBYTE,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/30756871/