Python 2.3.3 super() behaviour

[Nicolas Lehuen]

Ah, so there *is* more than one way to do it, then ? ;)

What the example shows is that super(...).__init__ does make one call to
each superclass' __init__, provided that those superclasses play nicely and
use super() themselves (as Peter wrote). If one of the superclasses does not
use super(), the 'magical' iteration over parent methods is interrupted,
hence the need to put those bad guys at the end of the superclasses list in
the class declaration.

But you're right, David, the simplest way to get what I want is to
explicitely call the superclasses' __init__. However, this would mean that
super() is not as useful as it seems. I'd rather find a way to *always* use
super() than have special cases for certain inheritance trees. In other
words, I'd rather have only one way to do it :).

Regards,
Nicolas

"David Fraser" <davidf at sjsoft.com> a écrit dans le message de
news:c65j4o$3me$1 at ctb-nnrp2.saix.net...
> super is not going to be helpful here, so why not call the X.__init__
> functions explicitly?
> Since you *need* multiple __init__ calls from the multiply-inherited
> class, super isn't going to work...
>
> Nicolas Lehuen wrote:
> > The only problem I have is that I want to build a multiple inheritance
> > involving an object which does not cooperate, namely :
> >
> > class T(object):
> >     def __init__(self):
> >         super(T,self).__init__()
> >
> > class  TL(list,object):
> >     def __init__(self)
> >         super(TL,self).__init__()
> >
> > In this case, T.__init__ is not called, because list.__init__ does not
use
> > super(). The only clean way to proceed is to change the inheritance
order :
> > TL(T,list). This way, both constructors are called.
> >
> > Here is another example which exhibits the behaviour :
> >
> > class A(object):
> >     def __init__(self):
> >         super(A,self).__init__()
> >         print 'A'
> >
> > class B(object):
> >     def __init__(self):
> >         print 'B'
> >
> > class C(B,A):
> >     def __init__(self):
> >         super(C,self).__init__()
> >         print 'C'
> >
> > class D(A,B):
> >     def __init__(self):
> >         super(D,self).__init__()
> >         print 'D'
> >
> >
> >>>>C()
> >
> > B
> > C
> > <__main__.C object at 0x008F3D70>
> >
> >>>>D()
> >
> > B
> > A
> > D
> > <__main__.D object at 0x008F39F0>
> >
> > The problem is that if you go further down the inheritance, the
behaviour is
> > even more complicated :
> >
> > class E(object):
> >     def __init__(self):
> >         super(E,self).__init__()
> >         print 'E'
> >
> > class F(C,E):
> >     def __init__(self):
> >         super(F,self).__init__()
> >         print 'F'
> >
> > class G(D,E):
> >     def __init__(self):
> >         super(G,self).__init__()
> >         print 'G'
> >
> >
> >>>>F()
> >
> > B
> > C
> > F
> > <__main__.F object at 0x008F3D70>
> >
> >>>>G()
> >
> > B
> > A
> > D
> > G
> > <__main__.G object at 0x008F3EF0>
> >
> > class H(E,C):
> >     def __init__(self):
> >         super(H,self).__init__()
> >         print 'H'
> >
> > class I(E,D):
> >     def __init__(self):
> >         super(I,self).__init__()
> >         print 'I'
> >
> >
> >>>>H()
> >
> > B
> > C
> > E
> > H
> > <__main__.H object at 0x008F3E30>
> >
> >>>>I()
> >
> > B
> > A
> > D
> > E
> > I
> > <__main__.I object at 0x008F3FD0>
> >
> > So the conclusion is : never do that :). Another more constructive
> > conclusion would be : always put the most cooperative classes first in
the
> > inheritance declaration, provided that it doesn't interfere with your
needs.
> > A class which has an uncooperative ancestor is less cooperative than a
class
> > which has only cooperative ancestors.
> >
> > Regards,
> > Nicolas
> >
> > "Nicolas Lehuen" <nicolas.lehuen at thecrmcompany.com> a écrit dans le
message
> > de news:40864674$0$24834$afc38c87 at news.easynet.fr...
> >
> >>OK, I get it now, thanks.
> >>
> >>super() method calls should only be used for method involved in
> >>diamond-shaped inheritance. This is logical since in this case the base
> >>classe (from which the diamond-shaped inheritance starts) defines the
> >>interface of the method.
> >>
> >>This solves another question I was asking myself about super() : "how
can
> >
> > it
> >
> >>work when the method signature differ between B and C ?". Answer : the
> >>method signature should not change because polymorphic calls would be
> >>greatly endangered. The base class defines the signature of the method
> >
> > which
> >
> >>must be followed by all its children, this way super() can work
properly.
> >>
> >>The base method signature is not enforced by Python, of course, but
you'd
> >>better respect it unless you get weird result in polymorphic calls.
> >>
> >>Regards,
> >>Nicolas
> >>
> >>"Peter Otten" <__peter__ at web.de> a écrit dans le message de
> >>news:c65fbo$1q4$05$1 at news.t-online.com...
> >>
> >>>Nicolas Lehuen wrote:
> >>>
> >>>
> >>>>Hi,
> >>>>
> >>>>I hope this is not a FAQ, but I have trouble understanding the
> >
> > behaviour
> >
> >>>>of the super() built-in function. I've read the excellent book 'Python
> >>
> >>in
> >>
> >>>>a Nutshell' which explains this built-in function on pages 89-90.
> >
> > Based
> >
> >>on
> >>
> >>>>the example on page 90, I wrote this test code :
> >>>>
> >>>>class A(object):
> >>>>    def test(self):
> >>>>        print 'A'
> >>>>
> >>>>class B(object):
> >>>>    def test(self):
> >>>>        print 'B'
> >>>>
> >>>>class C(A,B):
> >>>>    def test(self):
> >>>>        super(C,self).test()
> >>>>        print 'C'
> >>>>
> >>>>print C.__mro__
> >>>>c=C()
> >>>>c.test()
> >>>>
> >>>>The output is :
> >>>>(<class '__main__.C'>, <class '__main__.A'>, <class '__main__.B'>,
> >
> > <type
> >
> >>>>'object'>)
> >>>>A
> >>>>C
> >>>>
> >>>>Whereas I was expecting :
> >>>>(<class '__main__.C'>, <class '__main__.A'>, <class '__main__.B'>,
> >
> > <type
> >
> >>>>'object'>)
> >>>>A
> >>>>B
> >>>>C
> >>>>
> >>>>Was I wrong to expect this (based on what I've read ?)
> >>>
> >>>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()
> >>>
> >>>Here all cooperating methods have a super() call, and the base class
> >
> > acts
> >
> >>as
> >>
> >>>a showstopper to prevent that Python tries to invoke the non-existent
> >>>object.test().
> >>>
> >>>Peter
> >>>
> >>>
> >>>
> >>
> >>
> >
> >