[Python-Dev] [Python-3000] Pre-pre PEP for 'super' keyword

[Tim Delaney]

From: "Delaney, Timothy (Tim)" <tdelaney at avaya.com>

> Sorry - this is related to my proposal that the following two bits of
> code behave the same:
>
>    class A(object):
>        def f(self, *p, **kw):
>            super.f(*p, **kw)
>
>    class A(object):
>        def f(self, *p, **kw):
>            super(*p, **kw)
>
> But as has been pointed out, this creates an ambiguity with:
>
>    class A(object):
>        def f(self, *p, **kw):
>            super.__call__(*p, **kw)
>
> so I want to see if I can resolve it.

A 'super' instance would be callable, without being able to access it's 
__call__ method (because super.__call__ would refer to the base class method 
of that name).

But I find I really don't care. The only place where that would really 
matter IMO is if you want to find out if a 'super' instance is callable. 
Calling a base class __call__ method would not be ambiguous - the following 
two classes would work the same:

    class A(object):
        def __call__(self, *p, **kw):
            return super.__call__(*p, **kw)

    class A(object):
        def __call__(self, *p, **kw):
            return super(*p, **kw)

So, I guess my question is whether the most common case of calling the base 
class method with the same name is worth having some further syntactic sugar 
to avoid repetition? I think it is, but that would be your call Guido.

Cheers,

Tim Delaney