Pass template typename with free template argument

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?

score 6 · Accepted Answer · answered Jun 29 '12 at 16:33

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;
};

David Rodríguez - dribeas · Answer 2 · 2012-06-29T16:47:56.767

1

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;
};

edited Jun 29 '12 at 16:47

answered Jun 29 '12 at 16:33

David Rodríguez - dribeas

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I think that needs to be ``, otherwise e.g. `std::array` and `boost::array` wouldn't work. – ildjarn Jun 29 '12 at 16:34
@ildjarn: Yes, missed that. Thanks – David Rodríguez - dribeas Jun 29 '12 at 16:48

score 0 · Answer 3 · edited May 23 '17 at 12:21

0

If I understand what you're trying to accomplish, you can pass the template itself as a template template parameter.

edited May 23 '17 at 12:21

Community

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answered Jun 29 '12 at 16:31

Jerry Coffin

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That would be great! Can you explain? – user Jun 29 '12 at 16:32

Pass template typename with free template argument

3 Answers3