In article <mailman.3730.1372007386.3114.python-l...@python.org>, Ian Kelly <ian.g.ke...@gmail.com> wrote:
> Yes, you're missing that super() does not simply call the base class, > but rather the next class in the MRO for whatever the type of the > "self" argument is. If you write the above as: > > class Base1(object): > def __init__(self, foo, **kwargs): > super(Base1, self).__init__(**kwargs) > > class Base2(object): > def __init__(self, bar, **kwargs): > super(Base2, self).__init__(**kwargs) > > class Derived(Base1, Base2): > def __init__(self, **kwargs): > super(Derived, self).__init__(**kwargs) > > And then you create an instance of Derived by calling > Derived(foo='foo', bar='bar') and trace the call chain, you find that > Derived.__init__ will call Base1.__init__(foo='foo', bar='bar'), which > extracts its argument and then calls (surprise!) > Base2.__init__(bar='bar'), which again extracts its argument and then > calls object.__init__(), ending the chain. Mind. Blown. I'm tempted to go all Ranting Rick about this being non-obvious, but I see you already covered that in another post :-) The other thing I see here is that to make this work, the base classes need to accept **kwargs. That's kind of annoying, since it means a class has to explicitly designed to be multiply-inheritable. -- http://mail.python.org/mailman/listinfo/python-list
