问题
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)?
回答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.
来源：https://stackoverflow.com/questions/7136406/python-confused-by-super