关于< chrono>头在C ++ 11 [英] Several questions about <chrono> header in C++ 11

问题描述

我对C ++ 11中的新< chrono> 标题有几个问题。使用Windows 7，Visual Studio 2012。

查看示例 http://en.cppreference.com/w/cpp/chrono

  #include< iostream> 
 #include< chrono> 
 #include< ctime> 
 
 int fibonacci（int n）
 {
 if（n <3）return 1; 
 return fibonacci（n-1）+ fibonacci（n-2）; 
} 
 
 int main（）
 {
 std :: chrono :: time_point< std :: chrono :: system_clock>开始，结束
 start = std :: chrono :: system_clock :: now（）; 
 int result = fibonacci（42）; 
 end = std :: chrono :: system_clock :: now（）; 
 
 int elapsed_seconds = std :: chrono :: duration_cast< std :: chrono :: seconds> 
（end-start）.count（）; 
 std :: time_t end_time = std :: chrono :: system_clock :: to_time_t（end）; 
 
 std :: cout<< 完成计算在< std :: ctime（& end_time）
<< 经过时间：< elapsed_seconds<< s\\\
; 
} 
  

可能的输出

 已于6月16日完成计算20:42:57 2012 
已用时间：3s 
  

1. 我注意到该示例使用 std :: chrono :: system_clock :: now（）; 意味着它可以用来测量只流逝的时间，而不是CPU时间???如果我想测量CPU时间，我应该使用什么时钟？


2. 请注意，已用时间：3s 的输出会舍入为整数。

p>正确




根据标准：




system_clock表示[s]从系统范围的实时时钟到挂钟时间。




< chrono> library不提供测量CPU时间的机制，所以如果你想要你必须回到旧的< ctime> 库并使用 std :: clock（）。





（如果你的目标是Windows，你必须回到平台特定的API Windows提供的CPU时间，因为你指出，他们 std :: clock（）无法正常工作。





system_clock 更像是 std :: time（）的比较对象 std :: clock（）。 （例如，注意 system_clock 提供 system_clock :: time_point s和 time_t 。）我想象在< chrono> 中没有时钟用于测量CPU时间是由于标准委员会的时间限制，该功能比系统的挂钟和实时时钟少。





如果你想要CPU时间，但也希望< chrono> 提供，您应该实现符合标准中概述的时钟概念并提供CPU时间的时钟类型，可以使用 std :: clock（） / code>。




3. 的行
   
   

   
   

     int elapsed_seconds = std :: chrono :: duration_cast< std :: chrono :: seconds> 
   （end-start）.count（）; 
     

   
   

   是时间四舍五入为整数秒。您可以选择任何期限，也可以使用浮点数表示法，以允许非整数值：

     std :: int64_t elapsed_attoseconds = 
    std :: chrono :: duration_cast< std :: chrono :: duration< std :: int64_t，std :: atto>> 
   （end-start）.count（）; 
    
    double elapsed_seconds = 
    std :: chrono :: duration_cast< std :: chrono :: duration< double，std :: ratio< 1>> 
   （end-start）.count（）;请注意，在实际代码中，你应该避免使用 .count（）
   来逃避 chrono :: duration 提供的强类型，直到你绝对必须。
     auto total_duration = end  -  start; 
    auto seconds = std :: chrono :: duration_cast< std :: chrono :: seconds>（total_duration）; 
    auto milli = std :: chrono :: duration_cast< std :: chrono :: milliseconds>（total_duration  -  seconds）; 
    
    std :: cout<< seconds.count（）< s< milli.count（）<< ms\\\
   ; 
     
    







I have several questions about new <chrono> header in C++ 11. Using Windows 7, Visual Studio 2012.

Looking at the example http://en.cppreference.com/w/cpp/chrono

#include <iostream>
#include <chrono>
#include <ctime>

int fibonacci(int n)
{
    if (n < 3) return 1;
    return fibonacci(n-1) + fibonacci(n-2);
}

int main()
{
    std::chrono::time_point<std::chrono::system_clock> start, end;
    start = std::chrono::system_clock::now();
    int result = fibonacci(42);
    end = std::chrono::system_clock::now();

    int elapsed_seconds = std::chrono::duration_cast<std::chrono::seconds>
                             (end-start).count();
    std::time_t end_time = std::chrono::system_clock::to_time_t(end);

    std::cout << "finished computation at " << std::ctime(&end_time)
              << "elapsed time: " << elapsed_seconds << "s\n";
}

Possible output

finished computation at Sat Jun 16 20:42:57 2012
elapsed time: 3s

1. I have noticed that example uses std::chrono::system_clock::now(); does it mean it can be used to measure only elapsed time and not the CPU time ??? And if I want to measure CPU time, what Clocks should I use ?
2. Notice that elapsed time: 3s is output is rounded to whole integer. Is there way to make it more granulated?

解决方案

1. Correct

   According to the standard:

   system_clock represent[s] wall clock time from the system-wide realtime clock.

   The <chrono> library does not provide a mechanism for measuring CPU time, so if you want that you'll have to fall back on the old <ctime> library and use std::clock().

   (And if you're targeting Windows you'll have to fall back on whatever platform-specific API Windows provides for getting CPU time since, as you point out, their std::clock() doesn't work correctly.)

   system_clock is more like a counterpart to std::time() than to std::clock(). (E.g., note that system_clock provides conversions between system_clock::time_points and time_t.) I imagine that the lack of a clock in <chrono> for measuring CPU time is due to time constraints on the standard committee and the fact that that functionality is less used than the system's wall clock and real-time clocks.

   If you want CPU time but also want the benefits that <chrono> provides, you should implement a clock type that conforms to the Clock concept outlined in the standard and which provides CPU time, perhaps implemented internally using std::clock().

2. The line that says

   int elapsed_seconds = std::chrono::duration_cast<std::chrono::seconds>
                       (end-start).count();
   

   is what causes the time to be rounded to an integral number of seconds. You can choose any period you'd like, or you can use a floating point representation in order to allow non-integral values:

   std::int64_t elapsed_attoseconds =
       std::chrono::duration_cast<std::chrono::duration<std::int64_t, std::atto>>
           (end-start).count();
   
   double elapsed_seconds =
       std::chrono::duration_cast<std::chrono::duration<double,std::ratio<1>>>
           (end-start).count();
   

   Note that in real code you should avoid using .count() to escape the strong typing provided by chrono::duration until you absolutely must.

   auto total_duration = end - start;
   auto seconds = std::chrono::duration_cast<std::chrono::seconds>(total_duration);
   auto milli = std::chrono::duration_cast<std::chrono::milliseconds>(total_duration - seconds);
   
   std::cout << seconds.count() << "s " << milli.count() << "ms\n";
   

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