任意功能的定时器 [英] A timer for arbitrary functions

问题描述

我试图建立一个函数模板，可以测量任意类型的函数的执行时间。这是我到目前为止所尝试的：

I tried to build a function template that can measure the execution time of functions of arbitrary type. Here is what I've tried so far:

#include <chrono>
#include <iostream>
#include <type_traits>
#include <utility>

// Executes fn with arguments args and returns the time needed
// and the result of f if it is not void
template <class Fn, class... Args>
auto timer(Fn fn, Args... args)
    -> std::pair<double, decltype(fn(args...))> {
  static_assert(!std::is_void<decltype(fn(args...))>::value,
                "Call timer_void if return type is void!");
  auto start = std::chrono::high_resolution_clock::now();
  auto ret = fn(args...);
  auto end = std::chrono::high_resolution_clock::now();
  std::chrono::duration<double> elapsed_seconds = end - start;
  return { elapsed_seconds.count(), ret };
}

// If fn returns void, only the time is returned
template <class Fn, class... Args>
double timer_void(Fn fn, Args... args) {
  static_assert(std::is_void<decltype(fn(args...))>::value,
                "Call timer for non void return type");
  auto start = std::chrono::high_resolution_clock::now();
  fn(args...);
  auto end = std::chrono::high_resolution_clock::now();
  std::chrono::duration<double> elapsed_seconds = end - start;
  return elapsed_seconds.count();
}

int main () {
    //This call is ambigous if the templates have the same name
    std::cout << timer([](double a, double b){return a*b;},1,2).first;
}

注意，我必须有一个不同名称的函数 void（...）函数。有没有办法摆脱第二个函数？

Notice that I have to have a function with a different name for void(...) functions. Is there any way to get rid of the second function?

（这是我在第一个地方做的正确吗？）

(And is what I did correct in the first place?)

推荐答案

您可以使用 enable_if 或标签分派。 Enable_if 似乎是这种情况下更快的方式：

You can use enable_if or tag dispatching. Enable_if seems to be the quicker way in this case:

#include <type_traits>

template <class Fn, class... Args>
auto timer(Fn fn, Args && ... args) -> typename std::enable_if< 
    // First template argument is the enable condition
    !std::is_same< 
            decltype( fn( std::forward<Args>(args) ... )), 
            void >::value,
    // Second argument is the actual return type
    std::pair<double, decltype(fn(std::forward<Args>(args)...))> >::type
{
   // Implementation for the non-void case
}

template <class Fn, class... Args>
auto timer(Fn fn, Args &&... args) -> typename std::enable_if< 
    std::is_same< 
            decltype( fn( std::forward<Args>(args) ... )), 
            void >::value,
    double>::type
{
   // Implementation for void case
}

将参数传递给调用函数：

Also you should use perfect forwarding to pass the arguments to the called function:

 auto timer(Fn fn, Args && ... args) // ...
                      ~~~^   

当你调用函数时：

 auto ret = fn( std::forward<Args>(args)...);

演示。注意，这适用于函数，lambda和可调用对象;几乎所有的东西都有一个运算符（）。

Demo. Notice that this works with functions, lambda and callable objects; pretty much everything with an operator().

从设计的角度来看， code> std :: pair 。由于C ++ 11有 std :: tie ，返回对 / tuple 是从函数返回多个结果的合法方式。我会继续前进，并说，为了一致性在void case你应该返回一个元组只有一个元素。

From a design standpoint, I see no problem in returning a std::pair. Since C++11 has std::tie, returning a pair/ tuple is the legitimate way of returning multiple results from a function. I would go forward and say that for consistency in the void case you should return a tuple with only one element.

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