Initialize std::array of std::array

Question

I have and std::array of std::array, and say that I want to initialize all the arrays to {1,2,3}. We write:

std::array<std::array<int,3>,3> x{{1,2,3},{1,2,3},{1,2,3}};

This is not very handy. It becomes really messy when you have more than 3 arrays, or each array has more than 3 elements.

However, it becomes even worse if the size of the array is not known a priori:

template <size_t n, size_t T> struct foo{
  std::array<std::array<int,n>,T> x;
}

How can you initialize x? To make it clearer, I would like to initialize all the arrays in x to an array of a certain parameter that is given. That is, something like:

template <size_t n, size_t T> struct foo{
  static constexpr int N{20};

  std::array<std::array<int,n>,T> x;

  foo() : x{ {N,N,...}, {N,N,...}, ...} {}
}

(If that were possible). Any suggestion or ideas? I can always iterate through x and call the method fill, as in the following piece of code:

for (size_t idx = 0; idx < x[0].size(); idx++)
  x[idx].fill(N);

But that is not initialization, right? I am new to using std::array and I do not know whether I am asking something dummy here :/

Do you really need to initialize all the elements to a specific value? std::array<std::array<int,3>,3> x{} will make them all 0's if you are okay with that. — NathanOliver
– NathanOliver, Commented May 22, 2018 at 16:33
Does foo know what it should initialize the elements to? Should this be delegated to the caller? You're sort-of implying you know what the values should be, but don't actually say. — Useless
– Useless, Commented May 22, 2018 at 16:34
@NathanOliver, @Useless you are both correct. I know that all the elements should be initialized to some value other than 0. I will change the question. — enanone
– enanone, Commented May 22, 2018 at 16:36
@Holt The inner arrays consist on numerical values (be them int, bool, double,...). No tricks here ;) — enanone
– enanone, Commented May 22, 2018 at 16:56

Passer By · Accepted Answer · 2018-05-22 16:45:51Z

1

With std::index_sequence, you might do something like:

template <std::size_t ... Is, typename T>
constexpr std::array<T, sizeof...(Is)>
make_array(const T& value, std::index_sequence<Is...>)
{
    return {{(void(Is), value)...}};
}

template <std::size_t N, typename T>
constexpr std::array<T, N> make_array(const T& value)
{
    return make_array(value, std::make_index_sequence<N>());
}

and then:

template <size_t n, size_t T>
struct foo{
    static constexpr int N{20};

    std::array<std::array<int,n>,T> x;

    foo() : x{make_array<T>(make_array<n>(N))} {}
};

edited May 22, 2018 at 16:45

Passer By

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answered May 22, 2018 at 16:41

Jarod42

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3 Comments

enanone Over a year ago

I just tested it, it is insane (insanly good, I mean). Only issue here is that I need C++14 to make it work. Is there a way of fitting your solution to C++11? I am very thankful for your answer, though :)

Jarod42 Over a year ago

index_sequence might be implemented in C++11 too.

Jarod42 Over a year ago

For example: implementation-c14-make-integer-sequence

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