Forwarding variadic list of arguments

Question

Both of the following code compiles and performs as expected, are they different?

template<typename T, typename ...U>
auto time_function(T&& func, U&& ...args)
{
   std::cout << "timing" << std::endl;
   auto val = std::forward<T>(func)(std::forward<U...>(args...));
   std::cout << "timing over" << std::endl;
   return val;
}

template<typename T, typename ...U>
auto time_function(T&& func, U&& ...args)
{
   std::cout << "timing" << std::endl;
   auto val = std::forward<T>(func)(std::forward<U>(args)...);
   std::cout << "timing over" << std::endl;
   return val;
}

Looking at SO How would one call std::forward on all arguments in a variadic function?, second seems to be recommended, but doesn't the first do the same thing?

Answer 1

They are not the same. They are the same in the case that the arity of args is 1 or 0. otherwise it will fail to compile, consider..

#include <iostream>
using namespace std;
template<typename T, typename ...U>
auto time_function_1(T&& func, U&& ...args)
{

    std::cout<<"timing"<<std::endl;
    auto val = std::forward<T>(func)(std::forward<U...>(args...));
    std::cout<<"timing over"<<std::endl;
    return val;
}

    template<typename T, typename ...U>
auto time_function_2(T&& func, U&& ...args)
{

    std::cout<<"timing"<<std::endl;
    auto val = std::forward<T>(func)(std::forward<U>(args)...);
    std::cout<<"timing over"<<std::endl;
    return val;
}



int f (int){return 0;}

int y (int,int){return 0;}

int main() {
    time_function_1(f,1);
    time_function_2(f,1);

    time_function_1(y,1,2); // fail
    time_function_2(y,1,2);
    return 0;
}

Demo

for the failing case std::forward<U...>(args...) expands to forward<int, int>(int&, int&) and will fail to compile.

std::forward<U>(args)... expands to std::forward<int>(int&),std::forward<int>(int&)