Re: super() doesn't get superclass

On Thu, 20 Sep 2007 12:00:40 +1000, Ben Finney wrote:

In its latter form, it is worthless to me when I'm looking for "get
superclass of A", but its name and parameters and documentation all lead
me very strongly to believe otherwise.

Why are you looking for the superclass of A?

If it is specifically for the purpose of inheritance, then surely "which
class(es) is/are the superclass(es)" is an implementation detail that you
shouldn't care about?

In other words, in principle you want to do something like the following:

class MyClass(*base_classes):
def method(self, *args):
print args
return inherit_from_base_classes(self, 'method')(*args)
# could also be written as: self.__inherit__('method', *args)
# or even: self.__inherit__().method(*args)
# or similar.

The details of the inheritance are not important, so long as it calls the
right method of the right base-classes in the right order. You shouldn't
need to know what that order is (except to the extent you've defined the
base classes).

If that's what you want, then you don't need the class itself. You want
*something like* super(), even though the existing implementation of
super is sadly confusing and hard to use.

BUT I think, as far as I can tell, that super() does actually do the
right thing, *if* you can work out just what arguments to give it, and
provided all the base classes *and their bases classes* themselves also
call super().

If you actually want the super-class(es) themselves, heaven knows why,
then you can use MyClass.__base__ and MyClass.__bases__, although you
have to intuit this from communing with the cosmos, because dir(MyClass)
doesn't show them.



--
Steven.
.

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