[python]: confused by super()

Question

Possible Duplicate:
Understanding Python super()

Class B subclasses class A, so in B's __init__ we should call A's __init__ like this:

class B(A):
    def __init__(self):
        A.__init__(self)

But with super(), I saw something like this:

class B(A):
    def __init__(self):
        super(B, self).__init__()  #or super().__init__()

My questions are:

Why not super(B, self).__init__(self)? Just because the return proxy object is a bound one?
If I omit the second argument in super and the return proxy object is an unbound one, then should I write super(B).__init__(self)?

And many, many others -- please search first. [how-does-pythons-super-do-the-right-thing](http://stackoverflow.com/questions/607186/how-does-pythons-super-do-the-right-thing) [how-does-pythons-super-work-with-multiple-inheritance](http://stackoverflow.com/questions/3277367/how-does-pythons-super-work-with-multiple-inheritance) [usage-of-python-3-super](http://stackoverflow.com/questions/2771904/usage-of-python-3-super) — agf, Aug 21 '11 at 04:43

score 5 · Accepted Answer · answered Aug 21 '11 at 04:42

super() returns an instance of the base class, so self gets implicitly passed to __init__() like in any other method call.

With regards to your second question, that's correct. Calling super() without an instance as the second argument returns a reference to the class itself, not an instance constructed from your subclass instance.

[python]: confused by super()

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