c++ - 获取使用 std::chrono 花费的平均时间

Question

我有一个函数运行了超过一百万次。我想通过打印 10,000 次函数调用的持续时间总和来打印函数运行所需的持续时间。

在每个函数的开头我都有这样的东西:

int counter = 0;
auto duration_total = 0; //not sure about the type
std::chrono::high_resolution_clock::time_point t1, t2, duration;

t1 = std::chrono::high_resolution_clock::now(); 
Function f(){
  counter++;
}

t2 = std::chrono::high_resolution_clock::now();
duration= std::chrono::duration_cast<std::chrono::nanoseconds>( t2 - t1 ).count();
duration_total += duration;

if(counter %10000 == 0){
      long int average_duration = duration_total/10000;
      duration_total = 0;
      cout << average_duration << "\n";
}

我找不到添加持续时间然后获取其平均值的方法。

Answer 1

如果你看 std::chrono::duration<Rep,Period>::count , 你可以看到你可以使用

int duration = std::chrono::duration_cast<std::chrono::nanoseconds>( t2 - t1 ).count();

(或其他东西，例如 unsigned long )，因为返回值为

The number of ticks for this duration.

完整:

#include <iostream>
#include <chrono>

int main()
{
    int counter = 0;
    auto duration_total = 0; //not sure about the type
    std::chrono::high_resolution_clock::time_point t1, t2;

    t1 = std::chrono::high_resolution_clock::now(); 

    t2 = std::chrono::high_resolution_clock::now();
    int duration = std::chrono::duration_cast<std::chrono::nanoseconds>( t2 - t1 ).count();
    duration_total += duration;

    if(counter %10000 == 0){
          long int average_duration = duration_total/10000;
          duration_total = 0;
          std::cout << average_duration << "\n";
    }
}

参见 Coliru .