/* ex: set ts=2 et: */
/**
 *
 *  steven came into ##c and wanted to know if there was a way of "optimizing" the logic:
 *    if (a == -1 && b == -1) ...
 *
 *  268435456 iterations:
 *          neg1_obvious:  1.178 secs
 *           neg1_crafty:  1.179 secs
 *          neg1_bitwise:  1.009 secs
 *
 */


#include <stdio.h>
#include <limits.h>
#include <sys/time.h>

#define N_TIMES (1 << 28)

static void speed(const char *name, int (*f)(int, int))
{
  struct timeval tv[2];
  double d[2];
  unsigned n = N_TIMES;
  printf("%20s: ", name);
  gettimeofday(tv, NULL);
  while (n--)
    f(n, n);
  gettimeofday(tv + 1, NULL);
  d[0] = (tv[0].tv_sec * 1000000) + tv[0].tv_usec;
  d[1] = (tv[1].tv_sec * 1000000) + tv[1].tv_usec;
  printf(" %.3f secs\n", (d[1] - d[0]) / 1000000);
}

static int neg1_obvious(int a, int b)
{
  return -1 == a && -1 == b;
}

static int neg1_crafty(int a, int b)
{
  return -1 == a && a == b;
}

static int neg1_bitwise(int a, int b)
{
  return -1 == (-1 & (a & b));
}

int main(void)
{
  printf("%u iterations:\n", N_TIMES);
  speed("neg1_obvious", neg1_obvious);
  speed("neg1_crafty",  neg1_crafty);
  speed("neg1_bitwise", neg1_bitwise);
  return 0;
}