min()/max() implementation without using conditional branch instruction

minmax.c

/*
 * max()/min() implementation without any conditional branch instruction.
 * basic idea from:
 * https://twitter.com/vivisuke/status/539563588332310528
 * http://vivi.dyndns.org/blog/archives/306
 *
 * countermeasure against signed integer overflow by n-soda @ github
 *
 * NOTE:
 * this program is NOT truly portable,
 * because it is implementation-defined behavior how ">>" operator works
 * against a signed integer.
 *
 * This code does not use conditional branch with legacy CPUs which do not
 * have an instruction like x86 SETcc instruction, but for modern CPUs
 * which have a SETcc equivalent, the following way which is described in
 * http://graphics.stanford.edu/~seander/bithacks.html
 * is better:
 *	maximum: return x ^ ((x ^ y) & -(x < y));
 *	minimum: return y ^ ((x ^ y) & -(x < y));
 * and branch instructions don't have to be avoided with such legacy CPUs ;-)
 *
 */
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

#define INT_BIT		(sizeof(int) * CHAR_BIT)

int int_maximum(int a, int b)
{
	int diff = a - b - INT_MIN;
	int a_lt_b =
	    (((a - b) ^ ((a ^ b) & (a ^ (a - b)))) & (a | diff | -diff))
	    >> (INT_BIT - 1);
	return (a & ~a_lt_b) | (b & a_lt_b);
}

int int_minimum(int a, int b)
{
	int diff = a - b - INT_MIN;
	int a_lt_b =
	    (((a - b) ^ ((a ^ b) & (a ^ (a - b)))) & (a | diff | -diff))
	    >> (INT_BIT - 1);
	return (a & a_lt_b) | (b & ~a_lt_b);
}

int main(int argc, char **argv)
{
	int a, b;

	if (argc != 3) {
		fprintf(stderr, "Usage: %s <value-1> <value-2>\n", argv[0]);
		return 1;
	}
	a = atoi(argv[1]);
	b = atoi(argv[2]);

	printf("maximum(%d, %d) = %d\n", a, b, int_maximum(a, b));
	printf("minimum(%d, %d) = %d\n", a, b, int_minimum(a, b));
	return 0;
}