防止基准测试的空语句优化

Question

考虑以下代码:

#include <iostream>
#include <chrono>
#include <cmath>
#include <ctime>
#include <cstdlib>
template <class Counter, class Function, class... Args>
inline double benchmark(const Counter& counter, Function&& f, Args&&... args)
{
    const std::chrono::high_resolution_clock::time_point marker 
    = std::chrono::high_resolution_clock::now();
    for (Counter i = Counter(); i < counter; ++i) {
        f(args...);
    }
    return std::chrono::duration_cast<std::chrono::duration<double> >
    (std::chrono::high_resolution_clock::now()-marker).count();
}
int main(int argc, char* argv[])
{
   srand(time(nullptr));
   double y = rand()%10+1;
   std::cout<<benchmark(1000000, [](double x){return std::sin(x);}, y)<<"n";
   return 0;
}

函数benchmark测量函数的执行时间。问题是，在优化过程中，函数作为空语句被擦除。是否有办法强制函数真正执行?

编辑:1)我正在寻找一个解决方案在一个标准的c++(没有编译器特定的指令)2)如果f能够尽可能保持通用(例如void返回类型)，那就更好了

Answer 1

我发现这个解决方案使用一个易失性的临时:

#include <iostream>
#include <chrono>
#include <cmath>
template <class Clock = std::chrono::high_resolution_clock, class Counter, class Function, class... Args>
inline double benchmark(const Counter& counter, Function&& f, Args&&... args)
{
    volatile decltype(f(args...)) temporary = decltype(f(args...))();
    const typename Clock::time_point marker = Clock::now();
    for (Counter i = Counter(); i < counter; ++i) {
      temporary = f(args...);
    }
    return std::chrono::duration<double>(Clock::now()-marker).count();
}
int main(int argc, char* argv[])
{
   std::cout<<benchmark(1000000000, [](double x){return std::sin(x);}, 3.)<<"n";
   return 0;
}

如果您知道如何改进此代码，请注释

Answer 2

既然(匿名)函数返回一个值，为什么不在benchmark中捕获该值并使用它做一些微不足道的事情，例如将其添加到通过引用传递的值?像这样:

template <class Counter, class Function, class... Args>
inline double benchmark(double& sum, const Counter& counter, Function&& f, Args&&... args)
{
    const std::chrono::high_resolution_clock::time_point marker 
    = std::chrono::high_resolution_clock::now();
    for (Counter i = Counter(); i < counter; ++i) {
        sum += f(args...);
    }
    return std::chrono::duration_cast<std::chrono::duration<double> >
    (std::chrono::high_resolution_clock::now()-marker).count();
}

我认为编译器现在将很难优化函数调用(假设您打印或以某种方式使用main()中的总和)。