multi-dimensional array based on std::array

Question

I need a template that gives a multi-dimensional array based on std::array.

template <typename T, size_t...>
using MyArray = ? // -> here is something I don't know how to write...

The usage is as follows:

static_assert(std::is_same_v<
  MyArray<int, 2>, std::array<int, 2>
>);

static_assert(std::is_same_v<
  MyArray<int, 2, 3>, std::array<std::array<int, 3>, 2>
>);  // the dimension should be the opposite order

MyArray a;    // a stacked 2-D array of size 2x3
a[0][2] = 2;  // valid index

Any help will be appreciated!

Do you really need a multi-dimensional array? You can often use a m * n sized array instead of m instances of an n-sized array. Then you can emulate the two-dimensional behaviour with simple accessor and mutator functions. — tadman
– tadman, Commented Aug 9, 2018 at 17:40
I think this one is a duplicate of stackoverflow.com/questions/17759757/multidimensional-stdarray — goutnet
– goutnet, Commented Aug 9, 2018 at 17:49

max66 · Accepted Answer · 2018-08-09 19:31:39Z

4

I don't know how to make it with only a using; the best I can imagine needs the helps of an helper struct.

Something as follows

template <typename, std::size_t ...>
struct MyArrayHelper;

template <typename T, std::size_t D0, std::size_t ... Ds>
struct MyArrayHelper<T, D0, Ds...>
 { using type = std::array<typename MyArrayHelper<T, Ds...>::type, D0>; };

template <typename T>
struct MyArrayHelper<T>
 { using type = T; };

template <typename T, std::size_t ... Ds>
using MyArray = typename MyArrayHelper<T, Ds...>::type;

answered Aug 9, 2018 at 19:31

max66

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