Thread: C++ equivalent of super

Originally Posted by @nthony

@grumpy
I meant this by reusable/"resistant to change":

Code:

public class Foo extends Bar
{
    public void draw() { super.draw(); }
}

Yes, I knew what you meant.

Originally Posted by @nthony

But I've figured out a close approximation using a simple macro:

Code:

#define SUPER base
class Foo : public base {
..
}
undef SUPER

Your approach also specifically requires that all member functions of your class are implemented inline with the class declaration.

The typedef approach I suggested does what you want (with the exception that it is necessary to use "super::draw()" rather than "super.draw()"). That is a syntactic difference, not a functional one. If you insist on Java syntax, then use Java.

Originally Posted by @nthony

I agree providing hooks is more to do with creating interfaces (as is templating); however I'd tend to think that interfacing and derivation are separate concepts in OO design. Meaning one shouldn't have to interface to derive. (or rather, how would you derive from an interface in such a model?)

With that comment, you've shown a fundamental misunderstanding of OO design.

Interface is a fundamental requirement for derivation: the intent of derivation is that the derived class implements the same interface as the base class, but can provide different implementation of the behaviour required by that interface. Yes, with derivation, it is possible to extend an interface (eg a derived class provides the base class interface, plus more) but the key premise of derivation is still that the derived class supports the base class interface.