Should std::array have move constructor?

Question

Moving can't be implemented efficiently (O(1)) on std::array, so why does it have move constructor ?

Answer 1

std::array has a compiler generated move constructor, which allows all the elements of one instance to be moved into another. This is handy if the elements are efficiently moveable or if they are only movable:

#include <array>
#include <iostream>

struct Foo
{
  Foo()=default;
  Foo(Foo&&)
  {
    std::cout << "Foo(Foo&&)\n";
  }
  Foo& operator=(Foo&&)
  {
    std::cout << "operator=(Foo&&)\n";
    return *this;
  }
};

int main()
{
  std::array<Foo, 10> a;
  std::array<Foo, 10> b = std::move(a);
}

So I would say std::array should have a move copy constructor, specially since it comes for free. Not to have one would require for it to be actively disabled, and I cannot see any benefit in that.

Answer 2

To summarize and expand on other answers, array<T> should be moveable (when T itself is moveable) because:

• T may be efficiently moveable.
• T may be move-only.