Why does a move constructor require a default constructor for its members?

Question

I was trying to implement a move constructor for a class without a copy constructor. I got an error that the default constructor for a member of the class was missing.

Here's a trivial example to illustrate this:

struct A {
public:
        A() = delete;
        A(A const&) = delete;
        A(A &&a) {}
};

struct B {
        A a;
        B() = delete;
        B(B const&) = delete;
        B(B &&b) {}
};

Trying to compile this, I get:

move_without_default.cc: In constructor ‘B::B(B&&)’:
move_without_default.cc:15:11: error: use of deleted function ‘A::A()’
  B(B &&b) {}
           ^
move_without_default.cc:6:2: note: declared here
  A() = delete;
  ^

Why is this an error? Any way around it?

Answer 1

Why does a move constructor requires a default constructor for its members?

The move constructor that you defined default-constructs a member. If you default construct any members, then the default constructor is required for those members.

A constructor (be it regular, copy or move) default initializes the members that are not listed in the member initialization list nor have a default member initialization. B::a is not in the member initialization list of the move constructor (it doesn't have an initialization list at all) and it has no default member initialization.

Any way around it?

Most simply, use the default move constructor:

B(B&&) = default;

The default move constructor move-constructs the members.

Answer 2

Use the constructor's initializer list to initialize the A member. As written, the move constructor uses, as the compiler says, the default constructor for A.

B(B&& b) : a(std::move(b.a)) {}