Is it possible to extract array size from a template argument?

Question

If this is a duplicate I apologize. I looked around and found similar issues but nothing exactly like this.

If I instantiate a template like so...

MyClass<int[10]> c;

How can I write the template to get access to both the type and the array size? I've tried everything I can think of and I can't get it.

I was inspired by the std::function template which lets you use similar syntax to the function prototype, like...

std::function<int(MyClass&)> myfunc;

So I thought it would be nice to have something similar for the array and its size. I can use any of the newest c++ features (c++ 11/14).

Answer 1

You can add a partial specialization which looks like this:

template <typename T, ptrdiff_t N>
class MyClass<T[N]>
{
};

Here's a demo.

Answer 2

template<class Arr>
struct array_size {};
template<class T, size_t N>
struct array_size<T[N]>:std::integral_constant<std::size_t, N>{};
template<class Arr>
struct array_element {};
template<class Arr>
using array_element_t = typename array_element<Arr>::type;
template<class T, size_t N>
struct array_element<T[N]>{using type=T;};

now you can array_size<ArrType>{} and array_element_t<ArrType> without unpacking the type.

Answer 3

template <typename T, typename = void>
struct deduce
{
};

template <typename T>
struct deduce<T,
  typename ::std::enable_if<
    ::std::is_array<T>{}
  >::type
>
{
  using value_type =
    typename ::std::decay<decltype(::std::declval<T>()[0])>::type;

  static constexpr auto size = sizeof(T) / sizeof(value_type);
};