C/C++ Check to see if bit 31 is set in an unsigned int

Question

I'm trying to check and see if a bit is set in an unsigned int. I'm not sure how I can do this, but I assume it would be something like this. I'm trying to make the cdq instruction in C++ (but a function)

Here is what I have

unsigned int cdq(unsigned int eax)
{
     unsigned int edx = 0;

     if( (eax >> 31) & 1 ) { edx = 0xFFFFFFFF; }
     return edx
}

When I use the function with the following values:

cdq(0x12345678) bit 31 is set (1) so it should return (unsigned int)-1 cdq(0x01) bit 31 is not set (0) so it should return 0

The problem is it always returns 0, and I'm not sure why

Answer 1

cdq(0x12345678) bit 31 is set (1)

No, it's not ... the highest bit set is bit 28:

    1    2    3    4    5    6    7    8
 0001 0010 0011 0100 0101 0110 0111 1000
 ^  ^ 
 |  |
31  28

You code should work, but I would use

if( eax & (1U << 31) ) edx = 0xFFFFFFFF;

since it's a bit more direct and it shifts a constant rather than a variable so does less work at run time (although an optimizing compiler should produce the same code for both).

Actually I would write something like

int cdq(int eax)
{
    return eax < 0? -1 : 0;
}

By the way, your code doesn't actually implement cdq because your eax and edx variables are not the hardware eax and edx registers. And it's really not a very good idea to replicate ASM instructions as C functions anyway ... C has its own features for doing these sorts of things, e.g.,

int32_t foo = -0x12345678;
int64_t bar = (int64_t)foo;