Re: What does clock() calculate?

["Alex Vinokur" <alexvn at connect dot to>]

"Alex Vinokur" <alexvn@connect.to> wrote in message bmhbv9$fu1$2@sea.gmane.org">news:bmhbv9$fu1$2@sea.gmane.org...
> clock : "Calculates the best available approximation of the cumulative amount of
> time used by your program since it started." (From man clock).
>
> Does clock() calculate  user-used-time + system-used-time?
>

Yes, it is.

===========================================
Windows 2000 Professional
CYGWIN_NT-5.0 1.5.4(0.94/3/2)
gcc version 3.3.1 (cygming special)
===========================================

========= C code : File t.c : BEGIN =========

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <sys/resource.h>
#include <assert.h>

int main(int argc, char** argv)
{
clock_t  start_clock,
  end_clock;
struct rusage start_rusage,
  end_rusage;

clock_t diff_clock;

double diff_rusage_user,
 diff_rusage_system;

int i;

  assert (argc > 1);
  fprintf (stderr, "--------- Start ---------\n");

  start_clock = clock();
  getrusage(RUSAGE_SELF, &start_rusage);

  for (i = 0; i < atoi (argv[1]); i++) printf ("ABC\n");

  getrusage(RUSAGE_SELF, &end_rusage);
  end_clock = clock();

  assert (!(start_clock == (clock_t)(-1)));
  assert (!(end_clock == (clock_t)(-1)));


  diff_clock = end_clock - start_clock;
  fprintf (stderr, "\n");
  fprintf (stderr, "clock         : start = %lu ticks, end = %lu ticks\n", start_clock, end_clock);
  fprintf (stderr, "clock         : elapsed time = %.2f sec\n", (double)(diff_clock)/CLOCKS_PER_SEC);

  diff_rusage_user =
  (end_rusage.ru_utime.tv_sec - start_rusage.ru_utime.tv_sec)
  +
  (end_rusage.ru_utime.tv_usec - start_rusage.ru_utime.tv_usec)/ 1.0e6;
  fprintf (stderr, "\n");
  fprintf (stderr, "rusage user   : start = (%ld sec, %ld msec), end = (%ld sec, %ld msec)\n",
  start_rusage.ru_utime.tv_sec,
  start_rusage.ru_utime.tv_usec,
  end_rusage.ru_utime.tv_sec,
  end_rusage.ru_utime.tv_usec
  );
  fprintf (stderr, "rusage user   : elapsed time = %.2f sec\n", diff_rusage_user);


  diff_rusage_system =
  (end_rusage.ru_stime.tv_sec - start_rusage.ru_stime.tv_sec)
  +
  (end_rusage.ru_stime.tv_usec - start_rusage.ru_stime.tv_usec)/ 1.0e6;
  fprintf (stderr, "\n");
  fprintf (stderr, "rusage system : start = (%ld sec, %ld msec), end = (%ld sec, %ld msec)\n",
  start_rusage.ru_stime.tv_sec,
  start_rusage.ru_stime.tv_usec,
  end_rusage.ru_stime.tv_sec,
  end_rusage.ru_stime.tv_usec
  );
  fprintf (stderr, "rusage system : elapsed time = %.2f sec\n", diff_rusage_system);


  fprintf (stderr, "\n--------- Finish --------\n\n");

  return 0;
}

========= C code : File t.c : END ===========


========= Compilation & Run : BEGIN ===========

$ gcc t.c

$ a 10000

--------- Start ---------

ABC
ABC
ABC
[---omitted---]
ABC
ABC
ABC

clock         : start = 10 ticks, end = 900 ticks
clock         : elapsed time = 0.89 sec

rusage user   : start = (0 sec, 10000 msec), end = (0 sec, 540000 msec)
rusage user   : elapsed time = 0.53 sec

rusage system : start = (0 sec, 0 msec), end = (0 sec, 360000 msec)
rusage system : elapsed time = 0.36 sec

--------- Finish --------


========= Compilation & Run : END =============


We can see that
  clock elapsed time (0.89) == rusage user elapsed time (0.53) + rusage system elapsed time (0.36)


 =====================================
   Alex Vinokur
     mailto:alexvn@connect.to
     http://mathforum.org/library/view/10978.html
     news://news.gmane.org/gmane.comp.lang.c++.perfometer
   =====================================






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