Should the Copy-and-Swap Idiom become the Copy-and-Move Idiom in C++11?

Question

As explained in this answer, the copy-and-swap idiom is implemented as follows:

class MyClass
{
private:
    BigClass data;
    UnmovableClass *dataPtr;

pub         


        

Answer 1

It's been a long time since I asked this question, and I've known the answer for a while now, but I've put off writing the answer for it. Here it is.

The answer is no. The Copy-and-swap idiom should not become the Copy-and-move idiom.

An important part of Copy-and-swap (which is also Move-construct-and-swap) is a way to implement assignment operators with safe cleanup. The old data is swapped into a copy-constructed or move-constructed temporary. When the operation is done, the temporary is deleted, and its destructor is called.

The swap behaviour is there to be able to reuse the destructor, so you don't have to write any cleanup code in your assignment operators.

If there's no cleanup behaviour to be done and only assignment, then you should be able to declare the assignment operators as default and copy-and-swap isn't needed.

The move constructor itself usually doesn't require any clean-up behaviour, since it's a new object. The general simple approach is to make the move constructor invoke the default constructor, and then swap all the members with the move-from object. The moved-from object will then be like a bland default-constructed object.

However, in this question's observer pattern example, that's actually an exception where you have to do extra cleanup work because references to the old object need to be changed. In general, I would recommend making your observers and observables, and other design constructs based around references, unmovable whenever possible.