How to get duration, as int milli's and float seconds from <chrono>?

Question

I'm trying to use chrono library for timers and durations.

I want to be able to have a Duration frameStart; ( from app start ) and a Duration frameDelta; ( time between frames )

I need to be able to get the frameDelta duration as milliseconds and float seconds.

How do you do this with the new c++11 <chrono> libraries? I've been working on it and googling ( information is sparse ). The code is heavily templated and requires special casts and things, I can't figure out how to use this library correctly.

Answer 1

Is this what you're looking for?

#include <chrono>
#include <iostream>

int main()
{
    typedef std::chrono::high_resolution_clock Time;
    typedef std::chrono::milliseconds ms;
    typedef std::chrono::duration<float> fsec;
    auto t0 = Time::now();
    auto t1 = Time::now();
    fsec fs = t1 - t0;
    ms d = std::chrono::duration_cast<ms>(fs);
    std::cout << fs.count() << "s\n";
    std::cout << d.count() << "ms\n";
}

which for me prints out:

6.5e-08s
0ms

Answer 2

I don't know what "milliseconds and float seconds" means, but this should give you an idea:

#include <chrono>
#include <thread>
#include <iostream>

int main()
{
  auto then = std::chrono::system_clock::now();
  std::this_thread::sleep_for(std::chrono::seconds(1));
  auto now = std::chrono::system_clock::now();
  auto dur = now - then;
  typedef std::chrono::duration<float> float_seconds;
  auto secs = std::chrono::duration_cast<float_seconds>(dur);
  std::cout << secs.count() << '\n';
}

Answer 3

Taking a guess at what it is you're asking for. I'm assuming by millisecond frame timer you're looking for something that acts like the following,

double mticks()
{
    struct timeval tv;
    gettimeofday(&tv, 0);
    return (double) tv.tv_usec / 1000 + tv.tv_sec * 1000;
}

but uses std::chrono instead,

double mticks()
{
    typedef std::chrono::high_resolution_clock clock;
    typedef std::chrono::duration<float, std::milli> duration;

    static clock::time_point start = clock::now();
    duration elapsed = clock::now() - start;
    return elapsed.count();
}

Hope this helps.

Answer 4

In AAA style using the explicitly typed initializer idiom:

#include <chrono>
#include <iostream>

int main(){
  auto start = std::chrono::high_resolution_clock::now();
  // Code to time here...
  auto end = std::chrono::high_resolution_clock::now();

  auto dur = end - start;
  auto i_millis = std::chrono::duration_cast<std::chrono::milliseconds>(dur);
  auto f_secs = std::chrono::duration_cast<std::chrono::duration<float>>(dur);
  std::cout << i_millis.count() << '\n';
  std::cout << f_secs.count() << '\n';
}