How to Calculate Execution Time of a Code Snippet in C++

Question

Hi folks.

I should compute execution time of a C++ code snippet in seconds. It must be working either on Windows or Unix machines

I use code the following code to do this. (import before)

clock_t startTime = clock();
// some code here
// to compute its execution duration in runtime
cout << double( clock() - startTime ) / (double)CLOCKS_PER_SEC<< " seconds." << endl;

However for small inputs or short statements such as a = a + 1, I get "0 seconds" result. I think it must be something like 0.0000001 seconds or something like that.

I remember that System.nanoTime() in Java works pretty well in this case. However I can't get same exact functionality from clock() function of C++.

Do you have a solution?

Answer 1

Windows provides QueryPerformanceCounter() function, and Unix has gettimeofday() Both functions can measure at least 1 micro-second difference.

Answer 2

Try with this. It will work like time(NULL), but will return the number of milliseconds instead of seconds from the unix epoch on both windows and linux.

#ifdef WIN32
#include <Windows.h>
#else
#include <sys/time.h>
#include <ctime>
#endif

/* Returns the amount of milliseconds elapsed since the UNIX epoch. Works on both
 * windows and linux. */

int64 GetTimeMs64()
{
#ifdef WIN32
 /* Windows */
 FILETIME ft;
 LARGE_INTEGER li;

 /* Get the amount of 100 nano seconds intervals elapsed since January 1, 1601 (UTC) and copy it
  * to a LARGE_INTEGER structure. */
 GetSystemTimeAsFileTime(&ft);
 li.LowPart = ft.dwLowDateTime;
 li.HighPart = ft.dwHighDateTime;

 uint64 ret = li.QuadPart;
 ret -= 116444736000000000LL; /* Convert from file time to UNIX epoch time. */
 ret /= 10000; /* From 100 nano seconds (10^-7) to 1 millisecond (10^-3) intervals */

 return ret;
#else
 /* Linux */
 struct timeval tv;

 gettimeofday(&tv, NULL);

 uint64 ret = tv.tv_usec;
 /* Convert from micro seconds (10^-6) to milliseconds (10^-3) */
 ret /= 1000;

 /* Adds the seconds (10^0) after converting them to milliseconds (10^-3) */
 ret += (tv.tv_sec * 1000);

 return ret;
#endif
}

You can modify it to return microseconds instead of milliseconds if you want.

Answer 3

I have another working example that uses microseconds (UNIX, POSIX, etc).

    typedef unsigned long long timestamp_t;

    static timestamp_t
    get_timestamp ()
    {
      struct timeval now;
      gettimeofday (&now, NULL);
      return  now.tv_usec + (timestamp_t)now.tv_sec * 1000000;
    }

    ...
    timestamp_t t0 = get_timestamp();
    // Process
    timestamp_t t1 = get_timestamp();

    double secs = (t1 - t0) / 1000000.0L;

Here's the file where we coded this:

http://svn.arhuaco.org/svn/src/emqbit/tools/emqbit-bench/bench.c

Answer 4

I suggest using the standard library functions for obtaining time information from the system.

If you want finer resolution, perform more execution iterations. Instead of running the program once and obtaining samples, run it 1000 times or more.

Answer 5

It is better to run the inner loop several times with the performance timing only once and average by dividing inner loop repetitions than to run the whole thing (loop + performance timing) several times and average. This will reduce the overhead of the performance timing code vs your actual profiled section.

Wrap your timer calls for the appropriate system. For Windows, QueryPerformanceCounter is pretty fast and "safe" to use.

You can use "rdtsc" on any modern X86 PC as well but there may be issues on some multicore machines (core hopping may change timer) or if you have speed-step of some sort turned on.

Answer 6

In some programs I wrote I used RDTS for such purpose. RDTSC is not about time but about number of cycles from processor start. You have to etalonate it on your system to get a result in second, but it's really handy when you want to evaluate performance, it's even better to use number of cycles directly without trying to change them back to seconds.

(link above is to a french wikipedia page, but it has C++ code samples, english version is here)

Answer 7

boost::timer will probably give you as much accuracy as you'll need. It's nowhere near accurate enough to tell you how long a = a+1; will take, but I what reason would you have to time something that takes a couple nanoseconds?