Enforce static method overloading in child class in C++

Question

I have something like this:

class Base
{
  public:
    static int Lolz()
    {
      return 0;
    }
};

class Child : public Base
{
  public:
    int nothing;
};

template <typename T>
int Produce()
{
  return T::Lolz();
}

and

Produce<Base>();
Produce<Child>();

both return 0, which is of course correct, but unwanted. Is there anyway to enforce the explicit declaration of the Lolz() method in the second class, or maybe throwing an compile-time error when using Produce<Child>()?

Or is it bad OO design and I should do something completely different?

EDIT:

What I am basically trying to do, is to make something like this work:

Manager manager;

manager.RegisterProducer(&Woot::Produce, "Woot");
manager.RegisterProducer(&Goop::Produce, "Goop");

Object obj = manager.Produce("Woot");

or, more generally, an external abstract factory that doesn't know the types of objects it is producing, so that new types can be added without writing more code.

Answer 1

There are two ways to avoid it. Actually, it depends on what you want to say.

(1) Making Produce() as an interface of Base class.

template <typename T>
int Produce()
{
  return T::Lolz();
}
class Base
{
    friend int Produce<Base>();

protected:
    static int Lolz()
    {
        return 0;
    }
};

class Child : public Base
{
public:
    int nothing;
};

int main(void)
{
    Produce<Base>(); // Ok.
    Produce<Child>(); // error :'Base::Lolz' : cannot access protected member declared in class 'Base'
}

(2) Using template specialization.

template <typename T>
int Produce()
{
  return T::Lolz();
}
class Base
{
public:
    static int Lolz()
    {
        return 0;
    }
};

class Child : public Base
{
public:
    int nothing;
};

template<>
int Produce<Child>()
{
    throw std::bad_exception("oops!");
    return 0;
}

int main(void)
{
    Produce<Base>(); // Ok.
    Produce<Child>(); // it will throw an exception!
}