On Fri, Jan 16, 2009 at 2:12 PM, Terry Reedy <tjre...@udel.edu> wrote:
>
> I do not understand. You know it is going to run the .__init__ of its one
> and only base class, which here is object.
Because this class might be used as base of another class. Take this
trivial example code (in py2.6):
class A(object):
def __init__(self, a):
#super(A, self).__init__(a)
self.a = a
print "A"
class B(object):
def __init__(self, a):
#super(B, self).__init__(a)
self.b = a
print "B"
class C(A, B):
def __init__(self, a):
super(C, self).__init__(a)
self.c = a
print "C", dir(self)
C(1)
Running the last line shows that A's constructor got called, but not
B's constructor. The only way to make sure all __init__s are called in
this example is by doing
class A(object):
def __init__(self, a):
super(A, self).__init__(a)
self.a = a
print "A"
class B(object):
def __init__(self, a):
#super(B, self).__init__(a)
self.b = a
print "B"
class C(A, B):
def __init__(self, a):
super(C, self).__init__(a)
self.c = a
print "C", dir(self)
C(1)
which is really ugly (as in, why is B's call to super.__init__
commented but not A's, if A and B are otherwise identical?)
I'm not sure, but I think the proper behavior for object.__init__
should be ignoring all args.
--
- Alexandre
_______________________________________________
Python-Dev mailing list
Python-Dev@python.org
http://mail.python.org/mailman/listinfo/python-dev
Unsubscribe:
http://mail.python.org/mailman/options/python-dev/archive%40mail-archive.com
