Running time of functions

Question

I am wanting to print the running time of my functions. For some reason my timer always returns 0. Can anyone tell me why?

double RunningTime(clock_t time1, clock_t time2)
{
    double t=time1 - time2;
    double time = (t*1000)/CLOCKS_PER_SEC;
    return time;
}

int main()
{
     clock_t start_time = clock();


     // some code.....


    clock_t end_time = clock();

    std::cout << "Time elapsed: " << double(RunningTime(end_time, start_time)) << " ms";

    return 0;
}

I attempted to use gettimeofday and it still returned 0.

double get_time()
{
    struct timeval t;
    gettimeofday(&t, NULL);
    double d = t.tv_sec + (double) t.tv_usec/100000;
    return d;
}

int main()
{
        double time_start = get_time();

        //Some code......

        double time_end = get_time();

        std::cout << time_end - time_start;

    return 0;
}

Also tried using chrono and it gave me all kinds of build errors:

error: #error This file requires compiler and library support for the upcoming ISO C++ standard, C++0x. This support is currently
experimental, and must be enabled with the -std=c++0x or -std=gnu++0x compiler options.
warning: 'auto' will change meaning in C++0x; please remove it
error: ISO C++ forbids declaration of 't1' with no type error: 'std::chrono' has not been declared

error: request for member 'count' in '(t2 - t1)', which is of non-class type 'int'

int main() { auto t1 = std::chrono::high_resolution_clock::now();

            //Some code......

            auto t2 = std::chrono::high_resolution_clock::now();

            std::cout << "Time elapsed: " << std::chrono::duration_cast<std::chrono::milliseconds>(t2-t1).count() << " milliseconds\n";

        return 0;
    }

Consider using `` if you want good resolution. You can easily specify milliseconds for the units instead of calculating it as well. — chris, Jun 20 '12 at 00:10
On *nix systems, try `gettimeofday()` for high-resolution time (microseconds). — gavinb, Jun 20 '12 at 00:26
If you don't have C++11, you can consider `clock_gettime` on Linux (using `CLOCK_MONOTONIC_HR`), or `gethrtime` for most other UNIX variants, and `QueryPerformanceCounter` on Windows. — jxh, Jun 20 '12 at 00:27
@chris How would I implement `` ? I googled it real quick but can't quickly figure out how to put it in my program without causing build errors. I used `std::chrono::time_point time_point` — Jmh2013, Jun 20 '12 at 00:37
@Fourthmeal70, I see, you need the `-std=c++11` or `-std=c++0x` flag in your compiler options. — chris, Jun 20 '12 at 01:32
@chris thanks for that. I edited my code above to show the changes I made. I don't think my compiler agrees with `std::chrono` or I'm still using it wrong. My attempt at `chrono` is the last code block. — Jmh2013, Jun 20 '12 at 01:32
time taken to execute a complete function http://stackoverflow.com/a/40380118/6180077 — Abdullah Farweez, May 17 '17 at 05:04

Sergei Danielian · Accepted Answer · 2012-06-20T00:25:46.777

4

A timer tick is approximately equal to 1/CLOCKS_PER_SEC second, which is a millisecond resolution. To see a real (non-zero) number, you should either invoke a very long-time function or use another library with a higher time resolution facility:

new c++11x library chrono (use MSVS 2012)
boost::chrono (unfortunately, the library refers to a lot of others)
POSIX function gettimeofday, which gives you a 1 microsecond time resolution

edited Jun 20 '12 at 00:25

answered Jun 20 '12 at 00:20

Sergei Danielian

4,938
4
36
58

The weakness of `gettimeofday` is that it is not monotonic, and will generate strange results when something (like `ntpd`) adjusts the system time. – jxh Jun 20 '12 at 00:30
True, if a machine is attempting to keep synchronisation with a network time server, for example. But in this case it can be suited. – Sergei Danielian Jun 20 '12 at 00:33
Can you insert a 1 millisecond delay to your code and measure again? Or just repeat the code you want to measure for 100 times using loop, for example. I think your code performs very fast so you can't figure out how much time does it take to complete. – Sergei Danielian Jun 20 '12 at 00:51
What is the value of `timeval` structure? I mean both fields. – Sergei Danielian Jun 20 '12 at 00:56
I'm timing a sort function and used a file with 100,000 numbers and the timer says it took 0.312 ms. That seems short. I edited my code above to show all the ways I have tried it. – Jmh2013 Jun 20 '12 at 01:24
the values of `timeeval` : `tv_sec = 4686336` and `tv_usec = 4665580` – Jmh2013 Jun 20 '12 at 01:30
You were right. I was trying to time a function that was too fast. Thanks for your help. – Jmh2013 Jun 20 '12 at 01:52

score 0 · Answer 2 · answered Jun 20 '12 at 01:55

After lots of trial and error I went with gettimeofday. Here is my code that I finally got to work properly.

double get_time()
{
    struct timeval t;
    gettimeofday(&t, NULL);
    double d = t.tv_sec + (double) t.tv_usec/1000000;
    return d;
}

int main()
{
    double time_start = get_time();

    //Some code.........

    double time_end = get_time();

    std::cout << time_end - time_start;

    return 0;
}

score 0 · Answer 3 · answered Jan 05 '16 at 23:20

A solution I have been using lately uses C++11's lambda functionality to time any arbitrary function call or series of actions.

#include <ctime>
#include <iostream>
#include <functional>

void timeit(std::function<void()> func) {
    std::clock_t start = std::clock();

    func();

    int ms = (std::clock() - start) / (double) (CLOCKS_PER_SEC / 1000);

    std::cout << "Finished in " << ms << "ms" << std::endl;
}

int main() {
    timeit([] {
        for (int i = 0; i < 10; ++i) {
            std::cout << "i = " << i << std::endl;
        } 
    });

    return 0;
}

Running time of functions

3 Answers3

Linked

Related