Question

Possible Duplicate:
Accessing protected members in a derived class

If I have an abstract base class and a concrete templated class that derives from it, which has a method that uses a pointer to the base class – it seems that the derived class stops seeing itself as derived from it:

class AbstractBase
{
protected:
    virtual void test() = 0;
};

template < class T >
class Derived : public AbstractBase
{
public:
    virtual void call( AbstractBase* d ) { d->test(); }  //  Error!
protected:
    virtual void test() {}
};

int main()
{
    Derived< int > a;
    Derived< int > b;

    b.call( &a );

    return EXIT_SUCCESS;
}

This errors with:

‘virtual void AbstractBase::test()’ is protected

The compiler’s not wrong, it’s definitely protected – but if Derived< T > inherits from AbstractBase, why is it complaining?

Answer 1

The reason it isn’t allowed is because AbstractBase as a type declares test to be protected. This makes it private to all unless the current class is a direct descendant of AbstractBase. Even so, that class can only access the member though an object of the same class, not a different descendant, and not directly from AbstractBase itself.

template < class T >
class Derived : public AbstractBase
{
public:
    virtual void call( Derived * d ) {
        d->test(); // ok, d has same type as this
        AbstractBase *b = this;
        b->test(); // not ok
    }
protected:
    virtual void test() {}
};

You can, as demonstrated above, just allow it for pointers of the same type. Alternatively, you can create a proxy base class for Derived to implement your virtual method to call test. This will allow access from different Derived types.

class DerivedBase : public virtual AbstractBase
{
public:
    virtual void call( DerivedBase * d ) { d->test(); }
};

template < class T >
class Derived : public DerivedBase
{
protected:
    virtual void test() {}
};

And can be accessed this way:

   Derived< int > a;
   Derived< int > b;
   Derived< float > c;

   b.call( &a );
   c.call( &a );