多态性，为什么我不能这样做？

Question

class BaseA
{
}

class B : public BaseA
{
}

template <class T>
class C : public vector<T>
{
}

void someFunction (void)
{
    C<B> myClass;

    // I can't seem to do this...is it not possible?
    vector<BaseA> converted = ((vector<BaseA>) myClass);
}

See comment in code for what I am trying to do.

Answer 1

A vector of B isn't a vector of A even if B is an A (I assume a mixup between A and BaseA)

Try

vector<A> converted(myClass.begin(), myClass.end());

which is probably what you want to express.

(BTW inheriting from vector is bad idea in general , it isn't designed for that.)

Answer 2

C<B> and vector<A> are completely unrelated types and you would have to explicitly define such a conversion (eg template <typename TargetValueType> explicit operator vector<TargetValueType> () {... } ).

If such conversion is an uncommon task, something that is not really natural to the nature of your class, it might be preferable to use a range constructor (see AProgrammer's answer).

Also: If vector is std::vector , deriving from it is not a good idea. It isn't defined as a base class, and typically you should prefer containment over derivation (as it imposes looser coupling on your clients).

Answer 3

No way no how. Imagine if you had another class, X , derived from BaseA . If you could get your converted object, you could insert instances of X in it, even though the original object wouldn't allow it. You'd be breaking the semantics of your variables.

Incidentally you can do this in Java where they have no real sense of generics, but in a strongly typed generic language like C# or C++ you won't be able to do it (with the one exception in C#, but you shouldn't use it in this case anyway).