> > 1- > > 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...) after you telling me that dict.__init__(self...) should in your opinion the way to do it. I was not really happy because I did not understand at all this line. I start looking at this line for a while before understanding it (2 hours maybe)... :B
type(dict) <type 'type'> so dict is a class dict.__init__ is the function to initialize a dictionnary so dict.__init__(self...) initialize the dictionnary `self'. Which work because self derive from a dictionnary... I got it ! I was really lost because I was not clearly making a difference between the class: ``dict'' and an instance of it: ``dict()'' and thanks for your time. Cedric BRINER _______________________________________________ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
