From [12.8] [11] of N3337:
The implicitly-defined copy/move constructor for a non-union class X performs a memberwise copy/move of its bases and members. [ Note: brace-or-equal-initializers of non-static data members are ignored. See also the example in 12.6.2. —end note ] The order of initialization is the same as the order of initialization of bases and members in a user-defined constructor (see 12.6.2). Let
x
be either the parameter of the constructor or, for the move constructor, an xvalue referring to the parameter. Each base or non-static data member is copied/moved in the manner appropriate to its type:— if the member is an array, each element is direct-initialized with the corresponding subobject of x;
— if a member m has rvalue reference type T&&, it is direct-initialized with
static_cast<T&&>(x.m)
;— otherwise, the base or member is direct-initialized with the corresponding base or member of x.
This is really more of a clarification, but I can't see any mention of lvalue reference members in that clause. Since it doesn't mention them, by default it seems to say that they are part of the implicit memberwise move, yet the following would not work;
int x = 5;
int& y = x;
int& z(std::move(y)); //error: invalid initialization of non-const reference of type 'int&' from an rvalue of type 'std::remove_reference<int&>::type {aka int}'
So is it safe to assume that the default move constructor distinguishes that a member is a reference and would simply just do
int& z = y;
with no call to std::move
?
A move constructor allows the resources owned by an rvalue object to be moved into an lvalue without creating its copy.
A move constructor in general does not have to provide default initialization. Your move constructor does. A move constructor is still a constructor. And therefore, it must initialize all subobjects.
The move assignment operator is different than a move constructor because a move assignment operator is called on an existing object, while a move constructor is called on an object created by the operation. Thereafter, the other object's data is no longer valid.
In C++, the compiler creates a default constructor if we don't define our own constructor.
It's handled by the specification of class member access expressions. The key part is
Let
x
be either the parameter of the constructor or, for the move constructor, an xvalue referring to the parameter.
In other words, a defaulted move constructor for
struct X { int x, &y; };
does the equivalent of
X::X(X&& other) : x(std::move(other).x), y(std::move(other).y) {}
The important thing here is that the result of a class member access expression x.m
, where m
names a non-static data member, is always an lvalue if m
has reference type, but an xvalue if x
is an rvalue and m
has non-reference type. (See [expr.ref]/4.) This ensures that lvalue reference members will be initialized with lvalues.
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