Question

I have a template class for which I need to access a protected member function of the template parameter, like this:

class Foo
{
protected:
    void foo() {}
};

template<typename T>
class Bar
{
public:
    static void bar(T& self){self.foo();}
};
...
Foo f;
Bar<Foo>::bar(f);

My problem is getting access to the protected method. I tried putting a friend class T into Bar, but that doesn’t seem to be allowed in c++ (edit: and wouldn’t solve my problem anyways, so it seemd). I tried letting Bar inherit from T (template<typename T> class Bar: public T (could have used private inheritance, but the public interface of Bar is not terribly important, since the class itself is internal only)), but that didn’t allow for access of foo() either. So how do I get access to the foo() method?

Edit:
Foo should not need to know Bar<Foo>, since there are quite a lot Bar classes. I can however make other changes to Foo (without changing the public interface of course).

Answer 1

OK, this is a “rot in hell” hack. You can abuse the fact that you can form pointers-to-members pointing to protected base members from a derived class.

class Foo
{
protected:
    void foo() {}
};

// Helper template to bypass protected access control
// for a member function called foo, taking no parameters
// and returning void.
template<typename T>
struct Unprotect : public T
{
    typedef void (T::*FooPtr)();
    static FooPtr GetFooPtr()
    {
        return &Unprotect::foo;
    }
};

template<typename T>
class Bar
{
public:
    static void bar(T& self){(self.*Unprotect<Foo>::GetFooPtr())();}
};

int main()
{
    Foo f;
    Bar<Foo>::bar(f);
}