How to construct std::array object with initializer list? [duplicate]

Answer 1

Since a std::array is a structure that contains an aggregate (it is not an aggregate itself, and does not have a constructor that takes a std::initializer_list), you can initialize the underlying aggregate inside the structure with an initializer list using a double-braces syntax like so:

std::array<int, 4> my_array = {{1, 2, 3, 4}};

Note that this is not using std::initializer_list ... this is simply using a C++ initializer list to initialize the publicly accessible array member of std::array.

Answer 2

An std::array<> has no constructor that takes an std::initializer_list<> (initializer list constructor) and there is no special language support for what it may mean to pass a std::initializer_list<> to a class' constructors such that that may work. So that fails.

For it to work, your derived class needs to catch all elements and then forward them, a constructor template:

template<typename ...E>
enum_addressable_array(E&&...e) : base_t{{std::forward<E>(e)...}} {}

Note that you need {{...}} in this case because brace elision (omitting braces like in your case) does not work at that place. It's only allowed in declarations of the form T t = { ... }. Because an std::array<> consists of a struct embedding a raw array, that will need two level of braces. Unfortunately, I believe that the exact aggregate structure of std::array<> is unspecified, so you will need to hope that it works on most implementations.

Answer 3

An std::array does not have a constructor that takes an std::initializer_list. It's a good thing, because initializer lists can be bigger than the fixed-size of the array.

You can initialize it by testing that the initializer list is not larger than the size of the array and then copying the elements of initializer list to the elems member of std::array with std::copy.