是否为派生类生成了移动构造函数/赋值运算符

Question

If class X is derived from class Y and class Y has any of:

• A user-declared copy constructor,
• A user-declared copy assignment operator,
• A user-declared destructor
• A user-declared move constructor,
• A user-declared move assignment operator,

Will a move constructor and move assignment operator be implicitly defaulted for class X providing it declares none of the above?

eg

struct Y
{
     virtual ~Y() {}

     // .... stuff

};

struct X : public Y
{
   // ... stuff but no destructor, 
   //               no copy/move assignment operator 
   //               no copy/move constructor

   // will X have a default move constructor / assignment operator?
};

I am currently using gcc, but I am mainly interested in what the correct behaviour should be (as opposed to whether or not a particular compiler is standards compliant)..

Answer 1

As far as the derived class is concerned, the fact that an attribute or a base class has user-defined or implicit special functions does not matter. All that matter is their presence.

Compliant C++11 compilers should automatically provide the move constructors and assignment operators for the struct and class, if possible (which is clearly defined in the Standard), even though only classes with dynamically allocated buffers will really benefit from it (moving an int is just copying it).

Therefore, if your class embeds a std::string or a std::unique_ptr (for example), then its move constructor will call the embedded string or unique_ptr move constructor and it will be efficient... for free.

Simply changing the compilation mode should thus slightly improve performance.

Answer 2

Yes, they are, in §12.8.9:

If the definition of a class X does not explicitly declare a move constructor, one will be implicitly declared as defaulted if and only if

• X does not have a user-declared copy constructor,
• X does not have a user-declared copy assignment operator,
• X does not have a user-declared move assignment operator,
• X does not have a user-declared destructor, and
• the move constructor would not be implicitly defined as deleted.

and §12.8.10

The implicitly-declared move constructor for class X will have the form

X::X(X&&)

Similarly, for move assignment operators in 12.8.20:

If the definition of a class X does not explicitly declare a move assignment operator, one will be implicitly declared as defaulted if and only if

• X does not have a user-declared copy constructor,
• X does not have a user-declared move constructor,
• X does not have a user-declared copy assignment operator,
• X does not have a user-declared destructor, and
• the move assignment operator would not be implicitly defined as deleted.

Base classes don't come into it directly.