"Alex Vinokur" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] > 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:[EMAIL PROTECTED] http://mathforum.org/library/view/10978.html news://news.gmane.org/gmane.comp.lang.c++.perfometer ===================================== -- Unsubscribe info: http://cygwin.com/ml/#unsubscribe-simple Problem reports: http://cygwin.com/problems.html Documentation: http://cygwin.com/docs.html FAQ: http://cygwin.com/faq/