get non-type template parameters into tuple

Question

How does one construct a tuple with non type template parameters

template <auto... args>
void func()
{
  std::tuple<decltype(args)...> t(args...);
  cout << get<3>(t) << endl;
}

template <auto... args>
struct ZZ
{
  std::tuple<decltype(args)...> t(args...);
};


int main()
{
   func<1,2,3,4>();
   ZZ<1,2,3> z;
}

While it works for func it does not work for the struct and results in the compile error (gcc trunk)

vs.cc:102:35: error: ‘args’ is not a type
  102 |   std::tuple<decltype(args)...> t(args...);
      |                                   ^~~~

Answer 1

The problem is, the default member initializer (since C++11) supports braces and equal-sign initializer, but not parentheses initializer. You can change the code to:

template <auto... args>
struct ZZ
{
  std::tuple<decltype(args)...> t{args...};
  //                             ^       ^
};

Or

template <auto... args>
struct ZZ
{
  std::tuple<decltype(args)...> t = std::tuple<decltype(args)...>(args...);
  //                              ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
};

And with the help of class template argument deduction (since C++17):

template <auto... args>
struct ZZ
{
  std::tuple<decltype(args)...> t = std::tuple(args...);
  //                              ^^^^^^^^^^^^^^^^^^^^^
};