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:

1. Why not super(B, self).__init__(self)? Just because the return proxy object is a bound one?

2. 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)?

Answer 1

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.