DWORD to bytes using bitwise shift operators

Question

I can't get it to work correctly.

#include <windows.h>

int main()
{
    DWORD i  = 6521;

    BYTE first = i >> 32;
    BYTE second = i >> 24;
    BYTE third = i >> 16;
    BYTE fourth = i >> 8;

    i = (((DWORD)fourth) << 24) | (((DWORD)third) << 16) | (((DWORD)second) << 8) | first;
}

Answer 1

BYTE first = (i >> 24) & 0xff;
BYTE second = (i >> 16) & 0xff;
BYTE third = (i >> 8) & 0xff;
BYTE fourth = i & 0xff ;

Answer 2

I think You shift Your DWORD too much. By 8 bits too much :)

Answer 3

The bytes aren't always in the order that you expect, though Neil's solution is correct. You probably want to look at "endianess" if you're having that problem