this exemple will also works if you replace the: super(C,self).__init__( *args, **kw) by dict.__init__(self, *args, **kw)
but I do not understand this dict.__init_... call. Shouldn't you call the super class constructor??
super is just a convenience feature added to make Python slightly more like some other OOP languages. It is effectively just a wrapper around the explicit call to the super class:
Thus super(C,self...) is the same as
dict.__init__(self...)
No, super() is much smarter than that and was created to address deficiencies in direct superclass calling. super(C, self) actually finds the class that follows C in the method resolution order of the class of self. This can be very different from just calling the base class method; in the case of multiple inheritance super(C, self) may not ever refer to a base class of C.
For example this program:
class A(object):
def __init__(self, *args, **kwds):
print 'A.__init__()'
super(A, self).__init__(*args, **kwds)class B(object):
def __init__(self, *args, **kwds):
print 'B.__init__()'
super(B, self).__init__(*args, **kwds)class C(A, B):
def __init__(self, *args, **kwds):
print 'C.__init__()'
super(C, self).__init__(*args, **kwds)
C()prints: C.__init__() A.__init__() B.__init__()
For the original question (class SuperDict(dict,A)) this will not help because dict doesn't seem to call super(dict, self).__init__(). But if dict is listed as the last base class it works:
class mydict(A, dict):
def __init__(self, *args, **kwds):
print 'mydict.__init__()'
super(mydict, self).__init__(*args, **kwds)d=mydict(a=1, b=2) print d
prints:
mydict.__init__()
A.__init__()
{'a': 1, 'b': 2}so you can see both A.__init__() and dict.__init__() have been called.
For a fairly clear explanation see http://www.python.org/2.2/descrintro.html#cooperation
Kent
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