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;
}
CodePudding user response:
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;
}