C++ 获取 std::chrono::duration 的周期

Question

我在玩 std::chrono。当我做一些测试时，我想知道我是否可以获得用于构造 std::chrono::duration 的比率，因为我想打印它。

这里有一些代码可以显示我到底想做什么：

您应该能够通过添加 -std=c++11 标志在 windows 和 linux (g++) 上编译它。这个小示例代码应该测量你的机器需要的时间 cout to max int value.

main.cpp

#include<iostream>
#include "stopchrono.hpp"
#include<chrono>
#include<limit>

int main (){
    stopchrono<> main_timer(true);
    stopchrono<unsigned long long int,std::ratio<1,1000000000>,std::chrono::high_resolution_clock> m_timer(true);//<use long long int to store ticks,(1/1000000000)sekond per tick, obtain time_point from std::chrono::high_resolution_clock> 
    stopchrono<unsigned long long int,std::ratio<1,1000000000>> mtimer(true);


    std::cout<<"count to max of int ..."<<std::endl;
    for(int i=0;i<std::numeric_limits<int>::max();i++){}
    std::cout<<"finished."<<std::endl;

    main_timer.stop();
    m_timer.stop();
    mtimer.stop();
    std::cout<<std::endl<<"It took me "<<(main_timer.elapsed()).count()<<" Seconds."<<std::endl;
    std::cout<<"            "<<(m_timer.elapsed()).count()<<std::endl;//print amount of elapsed ticks by std::chrono::duration::count()
    std::cout<<"            "<<(mtimer.elapsed()).count()<<std::endl;

    std::cin.ignore();
    return 0;
}

stopchrono.hpp

#ifndef STOPCHRONO_DEFINED
#define STOPCHRONO_DEFINED

#include<chrono>

template<class rep=double,class period=std::ratio<1>,class clock=std::chrono::steady_clock> //this templates first two parameters determines the duration type that will be returned, the third parameter defines from which clock the duration will be obtained 
class stopchrono { // class for measurement of time programm parts are running
    typename clock::time_point start_point;
    std::chrono::duration<rep,period> elapsed_time;
    bool running;
public:
    stopchrono():
        start_point(clock::now()),
        elapsed_time(elapsed_time.zero()),
        running(false)
    {}
    stopchrono(bool runnit)://construct already started object
    running(runnit),
    elapsed_time(elapsed_time.zero()),
    start_point(clock::now())
    {}

    void start(){//set start_point to current clock::now() if not running
        if(!running){
            start_point=clock::now();
            running=true;
        }
    }
    void stop(){// add current duration to elapsed_time
        if(running){
            elapsed_time+=std::chrono::duration_cast<std::chrono::duration<rep,period>>(clock::now()-start_point);
            running=false;
        }
    }
    void reset(){// set elapsed_time to 0 and running to false
        elapsed_time=elapsed_time.zero();
        running=false;
    }
    std::chrono::duration<rep,period> elapsed(){//return elapsed_time
        if(running){
            return (std::chrono::duration_cast<std::chrono::duration<rep,period>>(elapsed_time+(clock::now()-start_point)));
        }else{
            return (elapsed_time);
        }
    }
    bool is_running()const{// determine if the timer is running
        return running;
    }
};
#endif

实际样本输出

count to max of int ...
finished.

It took me 81.6503 Seconds.
            81650329344
            81650331344

目标样本输出

count to max of int ...
finished.

It took me 81.6503 Seconds.
            81650329344 (1/1000000000)sekonds 
            81650331344

如何从返回的持续时间中获取已使用的时间段 std::ratio<1,1000000000>，即使我不知道我使用哪个时间段创建了秒表对象？这可能吗？

Answer 1

std::chrono::duration class 有一个 typedef period，这正是您要找的。您可以通过 decltype(your_variable)::period 访问它。像下面这样的事情应该做

auto elapsed = main_timer.elapsed();
cout << elapsed.count() << " " << decltype(elapsed)::period::num << "/"
     << decltype(elapsed)::period::den << endl;

另请参阅 this working example，它打印经过的时间和秒数比率。

C++ 获取 std::chrono::duration 的周期

C++ get period of an std::chrono::duration

c++

c++11

chrono