C++ Assign a variable to function call that could return void

Question

I'm trying to write a function that measures the time of execution of other functions. It should have the same return type as the measured function. The problem is that i'm getting a compiler error Variable has incomplete type 'void' when the return type is void. Is there a workaround to solve this problem? Help would be greatly appreciated, thanks!

#include <iostream>
#include <chrono>

template<class Func, typename... Parameters>
auto getTime(Func const &func, Parameters &&... args) {
    auto begin = std::chrono::system_clock::now();
    auto ret = func(std::forward<Parameters>(args)...);
    auto end = std::chrono::system_clock::now();
    std::cout << "The execution took " << std::chrono::duration<float>(end - begin).count() << " seconds.";
    return ret;
}

int a() { return 0; }
void b() {}

int main()
{
    getTime(a);
    getTime(b);
    return 0;
}

Answer 1

It's possible to solve this problem using specialization and an elaborate song-and-dance routine. But there's also a much simpler approach that takes advantage of return <void expression>; being allowed.

The trick is to fit it into this framework, by taking advantage of construction/destruction semantics.

#include <iostream>
#include <chrono>

struct measure_time {

    std::chrono::time_point<std::chrono::system_clock> begin=
        std::chrono::system_clock::now();

    ~measure_time()
    {
        auto end = std::chrono::system_clock::now();
        std::cout << "The execution took "
              << std::chrono::duration<float>(end - begin).count()
              << " seconds.\n";
    }
};


template<class Func, typename... Parameters>
auto getTime(Func const &func, Parameters &&... args) {

    measure_time measure_it;

    return func(std::forward<Parameters>(args)...);
}

int a() { return 0; }
void b() {}

int main()
{
    getTime(a);
    getTime(b);
    return 0;
}