传递带有免费模板参数的模板类型名

Question

In C++11 (or C++) is it possible to pass a template type that is not fully specified. Specifically, I want to pass a type that does not have all of its template specifiers defined yet:

template <std::size_t N, typename ARRAYTYPE>
struct A {
  ARRAYTYPE<int, N> int_array;
};

int main() {
  A<10, std::array> my_a;
  return 0;
}

I know that simply redefining ARRAYTYPE = std::array<int, 10> would work, but would not let me utilize an ARRAYTYPE of different size anywhere in A :

template <std::size_t N, typename ARRAYTYPE>
struct A {
  ARRAYTYPE<int, N> int_array;
  ARRAYTYPE<int, 1> tiny_int_array;
};

Is this possible?

Answer 1

It's called a "template template parameter", because it's a template parameter whose value is a template:

template <std::size_t N, template <typename, std::size_t> class ARRAYTYPE>
struct A {
  ARRAYTYPE<int, N> int_array;
  ARRAYTYPE<int, 1> tiny_int_array;
};

Answer 2

That would not be a type, but a template:

template <std::size_t N, template <typename, std::size_t> class ARRAY_TMPL>
struct A {
   ARRAY_TMPL<int, N> int_array;
};

Answer 3

如果我了解您要完成的工作，则可以将模板本身作为模板template参数传递。