Pergunta

Consider the following:

struct X
{
    Y y_;

    X(const Y & y) :y_(y) {}    
    X(Y && y) :y_(std::move(y)) {}
};

Is it necessary to define a constructor like the second one in order to take full advantage of move semantics? Or will it be taken care of automatically in the appropriate situations?

Foi útil?

Solução

Yes, but no. Your code should just be this:

struct X
{
    Y y_;

    X(Y y) : // either copy, move, or elide a Y
    y_(std::move(y)) // and move it to the member
    {} 
};

If you ever say in design "I need my own copy of this data"*, then you should just take the argument by value and move it to where it needs to be. It isn't your job to decide how to construct that value, that's up to the available constructors for that value, so let it make that choice, whatever it is, and work with the end result.

*This applies to functions too, of course, for example:

void add_to_map(std::string x, int y) // either copy, move or elide a std::string
{
    // and move it to where it needs to be
    someMap.insert(std::make_pair(std::move(x), y));
}

Note that is applied in C++03 too, somewhat, if a type was default constructible and swappable (which is all moving does anyway):

// C++03
struct X
{
    std::string y_;

    X(std::string y) // either copy or elide a std::string
    {
        swap(y_, y); // and "move" it to the member
    } 
};

Though this didn't seem to be as widely done.

Outras dicas

Yes, it is necessary. A const ref can only be a copy, not a move.

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