将右值引用分配给左值时发生了什么？

Question

I have a simple example:

void Message::move_Folders(Message *m) {
    m_folders = std::move(m->m_folders);
    for (auto f : m_folders) {
        f->removeMessage(m);
        f->addMessage(this);
    }
    m->m_folders.clear();
}

This function can move the "folders" member from one to another. So I want to know: Need I provide a move assign function for "folders"?

folders &operator=(folders &&f) {
    ...
}

What happened when assigning an rvalue reference to lvalue? Is it still a copy operation? (I believe not.)

Note: folders is a set of the Folder objects. A message may belong to many folders. So a message contains a set of Folders. The Folder also has a member messages . A folder may contain many messages. It's a little complicated.

This is the definition of Message and Folder :

class Message {
public:
// constructor and destructor
...
private:
    std::set<Folder *> m_folders;
...
};

class Folder {
friend class Message;
//constructor and destructor
...
private:
    std::set<Message *> m_messages;
};

Thanks

Answer 1

Note that std::move does not move a thing; rather it makes the thing movable . So in order to actually move the m->folders into this->folders , indeed the move assignment must perform that logic.

In fact, that move assignment should be the one responsible to define the move semantics: the move_folders method does not have to clear m->folders since it should not know this detail.

Answer 2

The need to provide a user defined move assignment depends on what other special members you have already defined, see the image for combinations:

So if Folders does not have a user defined destructor, copy constructor or assignment operator, then move constructor and assignment will be implicitly declared by the compiler. Otherwise you will need to write one yourself.

What happened when assigning an rvalue to lvalue? Is it still a copy operation? (I believe not.)

It depends if you have a move assignment it will move, else it will copy.

Finally do you need move assignment for your Folder class to move std::set is actually No, since the move assignment of std::set does not move the contents one by one with the default allocator.

Answer 3

You do not need to provide a move-assignment operator, but if you don't the call will indeed be downgraded to a copy.