Unexpected sign extension of int32 or 32bit pointer when converted to uint64

Question

I compiled this code using Visual Studio 2010 (cl.exe /W4) as a C file:

int main( int argc, char *argv[] )
{
    unsigned __int64 a = 0x00000000         


        

Answer 1

Use this to avoid the sign extension:

unsigned __int64 a = 0x00000000FFFFFFFFLL;

Note the L on the end. Without this it is interpreted as a 32-bit signed number (-1) and then cast.