Covariant return type and type conversion

Question

s->duplicate() returns an object of type Box*, but I\'m getting an error initializing it with Box*. It looks like it\'s being conve

Answer 1

Although Box::duplicate is being invoked at runtime (via virtual dispatch), and although Box::duplicate does override Shape::duplicate (covariantly), and although Box::duplicate does return a Box*, you'll still get a Shape* pointer because you are calling duplicate() through a Shape* pointer, and Shape* is the return type of Shape::duplicate(), and the compiler only sees you calling Shape::duplicate, not Box::duplicate.

C++ is not able to dynamically select types, so this is the best it can do. Your Box* is being automatically converted to a Shape* on the way out of Box::duplicate. As Barry said, "it still has to compile at compile time, and at compile time all we know is that it returns a Shape*".

Then, to make it into a Box* again, you need to explicitly cast it (using static_cast or dynamic_cast) because no implicit down-conversion exists.

[C++11: 10.3/7]: The return type of an overriding function shall be either identical to the return type of the overridden function or covariant with the classes of the functions. [..]

[C++11: 10.3/8]: If the return type of D::f differs from the return type of B::f, the class type in the return type of D::f shall be complete at the point of declaration of D::f or shall be the class type D. When the overriding function is called as the final overrider of the overridden function, its result is converted to the type returned by the (statically chosen) overridden function (5.2.2). [..]

In the standard text, a pertinent example follows.

Answer 2

The point isn't to do this:

Box*   b = s->duplicate();

That obviously can't work since Shape::duplicate() returns a Shape*. The point, rather, is to accept a Box* if you're calling duplicate() on a Box directly:

Box* old = new Box;
Box* b = old->duplicate(); // OK! We know it's a Box