Return 1 if any bits in an integer equal 1 using bit operations in C

Question

I've been thinking about this problem for hours. Here it is:

Write an expression that returns 1 if a given integer "x" has any bits equal to 1. return 0 otherwise.

I understand that I'm essentially just trying to figure out if x == 0 because that is the only int that has no 1 bits, but I can't figure out a solution. You may not use traditional control structures. You may use bitwise operators, addition, subtraction, and bit shifts. Suggestions?

Answer 1

Here's the best I could come up with:

y = (((-x) | x) >> (BITS - 1)) & 1;

where BITS = 32 for 32 bit ints, i.e. BITS = sizeof(int) * CHAR_BIT;

Here's a test program:

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

int main(int argc, char *argv[])
{
    const int BITS = sizeof(int) * CHAR_BIT;

    if (argc == 2)
    {
        int x = atoi(argv[1]);
        int y = (((-x) | x) >> (BITS - 1)) & 1;

        printf("%d -> %d\n", x, y);
    }

    return 0;
}