Python 2.3.3 super() behaviour

[Peter Otten]

Josef Meile wrote:

> Peter Otten wrote:
>  > As soon as a test() method without the super(...).test() is reached, no
>  > further test methods will be invoked. Only the first in the list of
>  > base classes will be invoked. If I'm getting it right you have to do
>  > something like:
>  >
>  > class Base(object):
>  >     def test(self):
>  >         print "base"
>  >
>  > class D1(Base):
>  >     def test(self):
>  >         super(D1, self).test()
>  >         print "derived 1"
>  >
>  > class D2(Base):
>  >     def test(self):
>  >         super(D2, self).test()
>  >         print "derived 2"
>  >
>  > class All(D1, D2):
>  >     pass
>  >
>  > All().test()
> Ok, this produces almost what the original poster wanted. You
> just have to invert the order of the base classes in All:
> 
>  >>>class All(D2, D1):
> ...   pass
> ...
> 
> then you will get:
>  >>> All().test()
> base
> derived 1
> derived 2
> 
> However, I don't understand jet why this doesn't print:
> 
> base
> derived 1
> base
> derived 2
> 
> The method test of Base is called just once? Why?
> I taught it was something like:
> 
> All.test() -> D1.test() + D2.test()
> 
> which is the same as:
> 
> (Base.test() + print "derived 1") + (Base.test() + print derived 2")
> 
> Thanks in advanced,
> Josef

Every instance has just one __dict__, so calling the same Base method twice
would at best be redundant. The Python designers are a bit smarter than
that and implemented "cooperative" methods for newstyle classes. See
http://www.python.org/2.2/descrintro.html#cooperation for the details.

Peter