How to create a multi dimensional std::array using a template?

user_9666018 0
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How can I create a multi dimensional std::array of type Type with (of course) known initial constexpr dimensions with some smart variadic template "MDA".

The number of dimensions shall be variable.

In the end I want to be able to write:

MDA<int,3,4,5,6> a4d{};

and the result should be equal to

std::array < std::array < std::array < std::array<int, 6> , 5 > , 4 > , 3 > a4d{};

And save a lot of (complicated, or even error prone) typing work . . .

Edit:

I am not sure if this possible at all. But what I am looking for, is some "typing saver", maybe in conjunction with a using statement.

variadic-templates templates arrays c++
2 Answers

Certainly possible with use of helper class templates, C++11 (except the static assert part)

#include <array>

template<typename T, std::size_t Head, std::size_t... Tail>
struct MDA_impl{
    using type = std::array<typename MDA_impl<T, Tail...>::type, Head>;
};
// Base specialization
template<typename T, std::size_t N>
struct MDA_impl<T,N>{
    using type = std::array<T, N>;
};

template<typename T, std::size_t... Ns>
using MDA = typename MDA_impl<T,Ns...>::type;

int main(){
    static_assert(std::is_same_v<MDA<int,3,4,5,6>,
                  std::array<std::array<std::array<std::array<int, 6>,5>,4>,3>>);
}
user_76917290

You can use the so-called meta-functions.

template<class T, int N, int... M>
struct MDA_struct
{
    using type = std::array<typename MDA_struct<T, M...>::type, N>;
};

template<class T, int N>
struct MDA_struct<T, N>
{
    using type = std::array<T, N>;
};

template<class T, int... N>
using MDA = typename MDA_struct<T, N...>::type;
user_152624890
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