Question

Whilst digging through the STL sources (DinkumWare, SGI, STLport, etc..) and trying to make sense of their implementation choices (it’s going well), I came across something I feel is a bit odd or rather ive never run into before.

Generally when one wishes to overload a member function in a derived class, you would prepend the base class member function signature with the virtual keyword, however at various points in the STL source this is not the case.

Here is a cut-down version of what I’m seeing in the STL implementations:

template <typename T> class A {
public:
    void func( ) { std::cout << "inside A func( )" << std::endl; }
};

template <typename T> class B : public A<T> {
public: 
    void func( ) { std::cout << "inside B func( )" << std::endl; }
};

The compiler seems fine with this pseudo-polymorphism, where as I was expecting an error something along the lines of:

error C2535: 'void B<T>::func(void)': member function already defined or declared

Would someone be kind enough to explain what is going on here?

PS: This also seems to work without the classes being templates too.

‘Regards

Answer 1

Without the virtual keyword – when redefining a function, you are hiding the super’s function.

In your case, by redifining func(), you tell the compiler there is a new function for B, which is different from A‘s.

Though, because it is not declared virtual, you will see this affect only if you invoke func() from a variable of type B. A variable of type A which holds a B, will invoke A‘s func().

A *a = new B;
a->func()

will invoke the first [A‘s] method.

To invoke B‘s method, you need the type to be B:

B *b = new B;
b->func()