Question

Possible Duplicate:
GCC left shift overflow

Consider the following minimal program.

#include <stdint.h>
#include <stdio.h>

int main()
{
    uint32_t v = 1024;
    v &= (((uint32_t)1 << 32) - 1);
    printf("v = %u\n", v);
    return 0;
}

This prints 1024 as I would expected compiling with GCC under MinGW. Because 1 shifted 32 times to the left is 0 again, thus 0-1 = -1 which is “1111….1111”. This AND’ed with any value should return the same value again.

However, if I change the program to

#include <stdint.h>
#include <stdio.h>

int main()
{
    unsigned int s = 32;
    uint32_t v = 1024;
    v &= (((uint32_t)1 << s) - 1);
    printf("v = %u\n", v);
    return 0;
}

The result printed is now 0. Can someone explain this behaviour?

Answer 1

Shifting a 32-bit value by 32 bits is undefined behaviour in C. Don’t do it.