将非类型模板参数放入元组

get non-type template parameters into tuple

如何构造带有非类型模板参数的元组

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;
}

虽然它适用于 func 但不适用于结构并导致编译错误 (gcc trunk)

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

问题是,default member initializer(C++11 起)支持大括号和等号初始化器,但不支持圆括号初始化器。您可以将代码更改为:

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

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

并借助 class template argument deduction (C++17 起):

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