You get this magical property of transforming an lvalue into an You might be wondering, how does one write a function like std::move? How do Prevent the compiler from providing an implicitly generated copyĬonstructor, and declaring a move assignment operator does not inhibit theĬreation of a standard assignment operator. On the other hand, declaring a move constructor does not The same is true here: addingĪ move constructor to a class will require you to declare and define your ownĭefault constructor. Longer generate the default constructor for you. Move constructors and implicitly generated constructorsĪs you know, in C++ when you declare any constructor, the compiler will no The same techniques as for creating a move constructor. Just as we have a move constructor, we should also have a : _name( std::move( other._name ) ) // oh, blissful efficiency
, _metadata( std::move( other._metadata ) )Īnd of course we should really go back to MetaData and fix its own move constructor so that it uses std::move on the string it holds: Move anything it just turns an lvalue into an rvalue, so that you can invoke Let's say that you have the following code:įor (auto itr = v.begin(), end_itr = v.end() itr != end_itr ++itr )įor ( int i = 0 i -std::move is a way of saying, "ok, honest to God I know I haveĪn lvalue, but I want it to be an rvalue." std::move does not, in and of itself, But this is notĪlways the case, and it can result in expensive object copies. The compiler (the return value optimization, for example). Sometimes these temporary objects can be optimized away by Unfortunately, until C++11, there hasīeen an obstinate wart that slows down many C++ programs: the creation of