Move constructors and `std::array`

Question

According to N3485 §23.3.2.2:

(...) the implicit move constructor and move assignment operator for array require that T be MoveConstructible or MoveAssignable, respectively.

So, std::array supports move semantics if the type of its elements does. Great!

However, what does this really mean? I tend to picture this type as a safer version of an array providing an STL-compliant interface but, if this is true, then how can an std::array move-construct its elements? Can I do the same with an ordinary array?

Answer 1

However, what does this really mean?

It means that, if the element type is movable, then so is the array type.

std::array<movable, 42> move_from = {...};
std::array<movable, 42> move_to = std::move(move_from); // moves all the elements

I tend to picture this type as a safer version of an array providing an STL-compliant interface

Not really. It's a wrapper for an array, giving it the same semantics as an aggregate class - including the ability to copy and move it.

how can an std::array move-construct its elements?

In exactly the same way as any other aggregate. Its implicit move-constructor will move-construct all its members, including the elements of any member arrays.

Can I do the same with an ordinary array?

Only if you wrap it in a class type, as std::array does.