how to print the time in linux

Question

so I was using rusage to print out how long it takes from the user, and system to process a command, something along the lines of,

 //DO STUFF HERE printf(" TOTAL TIMES: "); tusage.ru_utime.tv_sec = rusage.ru_utime.tv_sec + rusage.ru_stime.tv_sec;

@Art the same code prints 0.00 in linux, but same code prints actual value in mac os why is that? — NoNameY0
– NoNameY0, Commented Mar 14, 2013 at 8:07
@RichardMckenna because your program didn't take any measurable time to run? Your exact example works for me (even if I question why you'd use a struct rusage for storage for the sum instead of just a struct timeval). — Art
– Art, Commented Mar 14, 2013 at 8:09

perreal · Accepted Answer · 2013-03-14 08:15:23Z

Try this:

#include <stdio.h> #include <sys/resource.h> int main() { struct rusage rusage; struct rusage tusage; int i, j, r=0; for (i = 0; i < 10000; i++) { for (j = 1; j < 100000; j++) { r = i % j + i / j; } } getrusage(RUSAGE_SELF, &rusage); printf("TOTAL TIME \n"); tusage.ru_utime.tv_sec = rusage.ru_utime.tv_sec + rusage.ru_stime.tv_sec; tusage.ru_utime.tv_usec = rusage.ru_utime.tv_usec + rusage.ru_stime.tv_usec; tusage.ru_utime.tv_sec += tusage.ru_utime.tv_usec / 1000000; tusage.ru_utime.tv_usec = tusage.ru_utime.tv_usec % 1000000; printf("%ld.%06ld\n", tusage.ru_utime.tv_sec, tusage.ru_utime.tv_usec); return r; }

no it did not. But why does something on a mac take long yet all my realistic commands are showing up as 0.00
I don't really know, just a call to getrusage returns non-zero time for me (0.001000).

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