Using type traits with variadic template arguments

Question

I am attempting to make a print function that functions like a python print function, and I am having issues with enforcing what type the variadic arguments can be, I would like them to be restricted to const char*.

Here is my code so far:

#include <tuple>
#include <iostream>
#include <utility>
#include <type_traits>


template<
  std::size_t I=0,
  typename... Args,
  typename FUNCTION
  >
constexpr void for_each(const std::tuple<Args...>& t, FUNCTION &&func)
{
  if constexpr(I < sizeof...(Args))
  {
    func(std::get<I>(t));
    for_each<I + 1>(t, std::forward<FUNCTION>(func));
  }
}

template<
  typename... Args
  >
constexpr void print(Args... args)
{
  std::tuple t(std::forward<Args>(args)...);
  for_each(t, [=](const char* str)
  {
    std::cout << str << " ";
  });
  std::cout << '\n';
}

int main(void)
{
  print("this", "is", "a", "thing");
  print("this", "is", "not", "a", "thing");
  return 0;
}

I would like the variadic template argument in the print function to only accept const char*

Why would you limit yourself so? If you change [=](const char* str) to [=](const auto& elem ) your lambda can work with any type that has defined operator <<. — NathanOliver
– NathanOliver, Commented Jul 7, 2020 at 12:29

asmmo · Accepted Answer · 2020-07-07 12:41:29Z

5

You don't need std::tuple to do that and you can restrict your function to copy const char* only using std::enable_if, as follows

#include <iostream>
#include <type_traits>


template<typename ... Args>
constexpr std::enable_if_t<std::is_same_v<std::common_type_t<Args...>,const char*>, void>
print(Args&&... args)
{
    ((std::cout << args << " "),...);
    std::cout << '\n';
}
int main( )
{
    print("this", "is", "a", "thing");
    print("this", "is", "not", "a", "thing");
    print(5, "is", "not", "a", "thing");//compile error
    return 0;
}

edited Jul 7, 2020 at 12:41

answered Jul 7, 2020 at 12:36

asmmo

7,1401 gold badge14 silver badges27 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

finlay morrison Over a year ago

thanks for the fast answer, will this produce a substitution failure or will it just fail to compile?

asmmo Over a year ago

@finlaymorrison print(5, "is", "not", "a", "thing"); will prevent your compiler to compile the whole program

bartop · Accepted Answer · 2020-07-07 12:35:55Z

2

If you really want to accept only const char* you can do it like this, using static_assert (if you have C++17 support):

template<typename... Args>
constexpr void print(Args... args) {
  static_assert((std::is_same_v<Args, const char*> && ...), "Your pretty error message");
  std::tuple t(std::forward<Args>(args)...);
  for_each(t, [=](const char* str)
  {
    std::cout << str << " ";
  });
  std::cout << '\n';
}

answered Jul 7, 2020 at 12:35

bartop

10.5k1 gold badge28 silver badges66 bronze badges

4 Comments

finlay morrison Over a year ago

Thanks for the fast answer, i'll try implementing this.

Ted Lyngmo Over a year ago

What for_each overload is that? It looks wicked and nice. I found this implementation when looking for it.

bartop Over a year ago

@TedLyngmo its OP's for_each

Ted Lyngmo Over a year ago

@bartop Facepalm... I need to read more carefully :-) So, there - you also got a vote.

Collectives™ on Stack Overflow

Using type traits with variadic template arguments

2 Answers 2

2 Comments

4 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

4 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related