If I store a member as an object, will I incur an object copy during constuction?

Question

If the constructor for Door looks like this:

Door::Door(Doorknob doorknob) : m_doorknob(doorknob) { }

Then you would instantiate a Door like this:

Doorknob doorknob;
Door door(doorknob);  // Does an object copy of doorknob occur here?

It seems like if you store Doorknob as a pointer, you can explicitly avoid the copy:

Door::Door(Doorknob * doorknob_ptr) : m_doorknob_ptr(doorknob_ptr) { }

Instantiate Door like this:

Door door(new Doorknob);

But now you have to make sure to delete doorknob inside Door's destructor, which seems ugly.

What is the preferred approach?

Answer 1

Better approach is to pass with const reference:

Door::Door(const Doorknob& doorknob) : m_doorknob(doorknob) { }

Otherwise a copy of Doorknob will be passed to constructor and used as parameter for m_doorknob constructor.

Answer 2

More important than minimizing the number of copy constructor calls is that your code correctly models the problem space.

Door owns the doorknob; i.e. the lifecycle of DoorKnob is the same as that of Door. Thus, Door should manage the lifecycle of Doorknob.

The second solution is not be preferred for that reason. You are relying on the client to manufacture the Doorknob object, which you then store a pointer to, which is a violation of RAII.

Your first solution is acceptable, but as you note, as the problem of making an unnecessary copy of the Doorknob object, since copies as a parameter.

The ideal solution is to have contructor take in a reference, as @pingw33n's solution does. Or, since it may not make sense to construct a Doorknob outside of a door, have your Door constructor create the Door.