函数的运行时间 [英] Running time of functions

问题描述

我想打印我的功能的运行时间。由于某种原因，我的计时器总是返回0.任何人都可以告诉我为什么？

  double RunningTime（clock_t time1，clock_t time2）
 {
 double t = time1  -  time2; 
 double time =（t * 1000）/ CLOCKS_PER_SEC; 
 return time; 
} 
 
 int main（）
 {
 clock_t start_time = clock（）; 
 
 
 //一些代码..... 
 
 
 clock_t end_time = clock（）; 
 
 std :: cout<< 经过时间：< double（RunningTime（end_time，start_time））<  女士; 
 
 return 0; 
} 
  

我试图使用 gettimeofday 并返回0。

  double get_time（）
 {
 struct timeval t; 
 gettimeofday（& t，NULL）; 
 double d = t.tv_sec +（double）t.tv_usec / 100000; 
 return d; 
} 
 
 int main（）
 {
 double time_start = get_time（）; 
 
 //一些代码...... 
 
 double time_end = get_time（）; 
 
 std :: cout<< time_end  -  time_start; 
 
 return 0; 
} 
  

也尝试使用 chrono 它给我各种构建错误：

• 错误：#error此文件需要编译器和库支持
  即将到来的ISO C ++标准，C ++ 0x。此支持目前为
  
  实验性的，必须使用-std = c ++ 0x或-std = gnu ++ 0x
  编译器选项启用。


• warning：'auto'将改变C ++ 0x中的含义;请删除它


• 错误：ISO C ++禁止宣布't1'没有类型错误：
  'std :: chrono'尚未声明



• 错误：请求成员'count'在'（t2 - t1）'，这是
  非类型'int'

  
  
  

  int main（）
  {
  auto t1 = std :: chrono :: high_resolution_clock :: now（）;

    //有些代码...... 
   
   auto t2 = std :: chrono :: high_resolution_clock :: now（）; 
   
   std :: cout<< 经过时间：< std :: chrono :: duration_cast< std :: chrono :: milliseconds>（t2-t1）.count（）< milliseconds\\\
  ; 
   
   return 0; 
  } 
    

解决方案

计时器滴答大约等于1 / CLOCKS_PER_SEC秒，这是一个毫秒的分辨率。要查看一个实数（非零）数字，您应该调用一个很长时间的函数，或者使用另一个具有更高时间分辨率的库：

• new c ++ 11x library chrono （使用MSVS 2012）


• boost :: chrono （不幸的是，该库指的是很多其他人）


• POSIX函数 gettimeofday ，可提供1微秒的时间解决方案





I am wanting to print the running time of my functions. For some reason my timer always returns 0. Can anyone tell me why?

double RunningTime(clock_t time1, clock_t time2)
{
    double t=time1 - time2;
    double time = (t*1000)/CLOCKS_PER_SEC;
    return time;
}

int main()
{
     clock_t start_time = clock();


     // some code.....


    clock_t end_time = clock();

    std::cout << "Time elapsed: " << double(RunningTime(end_time, start_time)) << " ms";

    return 0;
}

I attempted to use gettimeofday and it still returned 0.

double get_time()
{
    struct timeval t;
    gettimeofday(&t, NULL);
    double d = t.tv_sec + (double) t.tv_usec/100000;
    return d;
}

int main()
{
        double time_start = get_time();

        //Some code......

        double time_end = get_time();

        std::cout << time_end - time_start;

    return 0;
}

Also tried using chrono and it gave me all kinds of build errors:

• error: #error This file requires compiler and library support for the upcoming ISO C++ standard, C++0x. This support is currently
  experimental, and must be enabled with the -std=c++0x or -std=gnu++0x compiler options.
• warning: 'auto' will change meaning in C++0x; please remove it
• error: ISO C++ forbids declaration of 't1' with no type error: 'std::chrono' has not been declared

• error: request for member 'count' in '(t2 - t1)', which is of non-class type 'int'

  int main() { auto t1 = std::chrono::high_resolution_clock::now();

              //Some code......
  
              auto t2 = std::chrono::high_resolution_clock::now();
  
              std::cout << "Time elapsed: " << std::chrono::duration_cast<std::chrono::milliseconds>(t2-t1).count() << " milliseconds\n";
  
          return 0;
      }
  

解决方案

A timer tick is approximately equal to 1/CLOCKS_PER_SEC second, which is a millisecond resolution. To see a real (non-zero) number, you should either invoke a very long-time function or use another library with a higher time resolution facility:

• new c++11x library chrono (use MSVS 2012)
• boost::chrono (unfortunately, the library refers to a lot of others)
• POSIX function gettimeofday, which gives you a 1 microsecond time resolution

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