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How to get elapsed time in seconds from boost::timer::cpu_timer?

Tags:

c++

boost

The following rough code, based on the documentation, gives me the elapsed time in seconds from the timer object provided in boost.

boost::timer::cpu_timer timer;
// ...do some work...
const boost::timer::nanosecond_type oneSecond(1000000000LL);
return timer.elapsed().user / oneSecond;

The problem with this method is that I have this uncomfortable magic number in my code. Is there some method within boost that can give me elapsed seconds out of the nanosecond_type value available from the call to elapsed().user, without having this magic number sitting in my code?

(EDIT:) Conclusion:

Based on the accepted answer, I ended up with this snippet in my production code:

boost::timer::cpu_timer timer;
// ...do some work...
auto nanoseconds = boost::chrono::nanoseconds(timer.elapsed().user + timer.elapsed().system);
auto seconds = boost::chrono::duration_cast<boost::chrono::seconds>(nanoseconds);
std::cout << seconds.count() << std::endl;
like image 871
Boinst Avatar asked Jun 26 '13 02:06

Boinst


2 Answers

As @jogojapan suggested, boost::chrono would be a good choice. But you don't need the duration_cast<> if you use double as underlying type.

typedef boost::chrono::duration<double> sec; // seconds, stored with a double
sec seconds = boost::chrono::nanoseconds(timer.elapsed().user);
std::cout << seconds.count() << "s\n"; // gives the number of seconds, as double.
like image 74
mr_georg Avatar answered Oct 05 '22 22:10

mr_georg


One more addition:

double
getSeconds( const boost::timer::nanosecond_type& elapsed )
{
  return
    static_cast< double >(
      boost::chrono::nanoseconds( elapsed ).count()
    ) * boost::chrono::nanoseconds::period::num / boost::chrono::nanoseconds::period::den;
}

This you can call with timer.elapsed().user for example to get a double value of the elapsed time. This can obviously also transformed to std::chrono, if wanted / supported.

like image 28
Arne Avatar answered Oct 05 '22 22:10

Arne