/*
 * longrun.c
 *
 *	This program runs for a very long time, and occasionally prints
 *	out messages that identify itself.
 *
 * Author: Ethan L. Miller (elm at cs.ucsc.edu)
 *
 * $Id: longrun.c,v 1.1 2006/05/02 17:23:29 elm Exp $
 */

#include <stdio.h>
#include <sys/types.h>
#include <unistd.h>

#define	LOOP_COUNT_MIN	100
#define	LOOP_COUNT_MAX	100000000
int
main (int argc, char *argv[])
{
  char *idStr;
  unsigned int	v;
  int	i = 0;
  int	iteration = 1;
  int	loopCount;
  int	maxloops;

  if (argc < 3 || argc > 4) {
    printf ("Usage: %s <id> <loop count> [max loops]\n", argv[0]);
    exit (-1);
  }
  /* Start with PID so result is unpredictable */
  v = getpid ();
  /* ID string is first argument */
  idStr = argv[1];
  /* loop count is second argument */
  loopCount = atoi (argv[2]);
  if ((loopCount < LOOP_COUNT_MIN) || (loopCount > LOOP_COUNT_MAX)) {
    printf ("%s: loop count must be between %d and %d (passed %d)\n",
	    argv[0], LOOP_COUNT_MIN, LOOP_COUNT_MAX, argv[2]);
    exit (-1);
  }
  /* max loops is third argument (if present) */
  if (argc == 4) {
    maxloops = atoi (argv[3]);
  } else {
    maxloops = 0;
  }

  /* Loop forever - use CTRL-C to exit the program */
  while (1) {
    /* This calculation is done to keep the value of v unpredictable.  Since
       the compiler can't calculate it in advance (even from the original
       value of v and the loop count), it has to do the loop. */
    v = (v << 4) - v;
    if (++i == loopCount) {
      /* Exit if we've reached the maximum number of loops.  If maxloops is
	 0 (or negative), this'll never happen... */
      if (iteration == maxloops) {
	break;
      }
      printf ("%s:%06d\n", idStr, iteration);
      fflush (stdout);
      iteration += 1;
      i = 0;
    }
  }
  /* Print a value for v that's unpredictable so the compiler can't
     optimize the loop away.  Note that this works because the compiler
     can't tell in advance that it's not an infinite loop. */
  printf ("The final value of v is 0x%08x\n", v);
}