How do I print out the arguments of a function using a variadic template?

Question

This example uses a common variadic template and function. I want to print out the arguments passed to f:

#include 

template <         


        

Answer 1

c++17 Updates!

Since the question is general and in C++17 you can do it better, I would like to give two approaches.

Solution - I

Using fold expression, this could be simply

#include 
#include   // std::forward

template
constexpr void print(Args&&... args) noexcept
{
   ((std::cout << std::forward(args) << " "), ...);
}

int main()
{
   print("foo", 10, 20.8, 'c', 4.04f);
}

output:

foo 10 20.8 c 4.04 

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---

Solution - II

With the help of if constexpr, now we can avoid providing base case/ 0-argument case to recursive variadic template function. This is because the compiler discards the false statement in the if constexpr at compile time.

#include 
#include   // std::forward

template 
constexpr void print(T&& first, Ts&&... rest) noexcept
{
   if constexpr (sizeof...(Ts) == 0)
   {
      std::cout << first;               // for only 1-arguments
   }
   else
   {
      std::cout << first << " ";        // print the 1 argument
      print(std::forward(rest)...); // pass the rest further
   }
}

int main()
{
   print("foo", 10, 20.8, 'c', 4.04f);
}

output

foo 10 20.8 c 4.04

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