Timing a function in a C++ program that runs on Linux

Question

I am trying to time a function of my C++ program using the amount of time that it takes to execute in user space. I tried the clock_gettime(CLOCK_PROCESS_CPUTIME_ID, &start) command from inside the program but I am afraid that this is the CPU time and not the User Time that I actually need. The time "program name" will not work in this case because I am only timing a function. Any help would be great. Thanks!

Have you considered using a profiler like gprof?

linuxuser27
– linuxuser27

2011-02-19 20:57:35 +00:00
Commented Feb 19, 2011 at 20:57 — linuxuser27
– linuxuser27, Commented Feb 19, 2011 at 20:57
what is gprof, and how do i use it?

Joe Shmo
– Joe Shmo

2011-02-20 23:09:55 +00:00
Commented Feb 20, 2011 at 23:09 — Joe Shmo
– Joe Shmo, Commented Feb 20, 2011 at 23:09

payne · Accepted Answer · 2011-02-19 21:25:34Z

5

Use the times() function, see: http://linux.die.net/man/2/times

This will give you current user and system time for your process. You can call it on entry and exit from your subroutine, and subtract.

You can also use a profiler like gprof, which will do this all automatically for you (but will incur overhead).

edited Feb 19, 2011 at 21:25

answered Feb 19, 2011 at 21:09

payne

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3 Comments

6502 Over a year ago

+1: times seems the perfect solution. The overhead of gprof isn't IMO that little and moreover instrumented code can behave quite differently for caching issues. Especially for fast code IMO the only way to get reasonable numbers are passive statistical profilers.

Joe Shmo Over a year ago

when I use times() and get the user time I get a value of 0. Doesnt that mean that my function is running too fast for it to time the function?

payne Over a year ago

That may be the case. times() returns a count of "clock ticks", and you have to call sysconf(_SC_CLK_TCK) to figure out what the tick period is on your system. See the times man page.

Community · Accepted Answer · 2017-05-23 12:31:02Z

You could use gettimeofday() as exemplified here.

Add this function to your code:

#include <sys/time.h> long myclock() { struct timeval tv; gettimeofday(&tv, NULL); return (tv.tv_sec * 1000000) + tv.tv_usec; }

and use it to retrieve the time:

long start = myclock(); // do something // ... long end = myclock() - start; std::cout << "[" << time->tm_hour << ":"<< time->tm_min << ":" << time->tm_sec << "] time:" << std::setprecision(3) << end/1000000.0 << std::endl;

Doesn't that measure wall-clock time rather than user-space CPU time? Your approach is what I generally use, but it is slightly different from the question seems to be asking. It also isn't clear if the result required is 'only this function' or 'this function and those it calls'.
The main problem on a system with hundreds of processes running is that you cannot know whether you were swapped out. So this method will really only work if your function is very fast and somehow your process doesn't get swapped for another while timing your function.

Sameer · Accepted Answer · 2013-06-25 06:36:31Z

Try the boost:timer library. It is portable and easy to use. The Boost timers separately give you wall clock time, user time and system time.

http://www.boost.org/doc/libs/1_53_0/libs/timer/doc/cpu_timers.html

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