Way to call super(MyClass, self).__init__() without MyClass?

Question

I find this syntax astoundingly annoying. Every time I rename my class, I have to change this call for no apparent reason. Isn't there some __class__ magic variable or something I can use at least? Interested in answers for Python 2.5, but it doesn't hurt to know if later versions fixed this.

Answer 1

As far as I know, this isn't possible in 2.5. However, in 3.0, this was changed so that you can simply call super().__init__().

Answer 2

In Python 3.0, the super() can be called without arguments to do the same thing.

Answer 3

This is fixed in Python 3. http://docs.python.org/py3k/library/functions.html#super

http://www.python.org/dev/peps/pep-3135/

Answer 4

I think if your class only inherits from one class it is safe to do just this:

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

But I could be mistaken.

Answer 5

EDIT

As pointed out by Alex, this is wrong where there is more than a single level of inheritence.

Yes, "new" style classes have a __class__ attribute available which can be used, eg.

class B(object):
    def __init__(self):
        print "B.__init__():"

class D(B):
    def __init__(self):
        print "D.__init__():"
        super(self.__class__, self).__init__()

>>> d = D()
D.__init__():
B.__init__():

>>> dir(d)
['__class__', '__delattr__', '__dict__', '__doc__', '__getattribute__', '__hash__', '__init__', '__module__', '__new__', '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__str__', '__weakref__']
>>> d.__class__
<class '__main__.D'>

Answer 6

There is some __class__ magic variable. The __class__ magic variable. Try:

class foo(OtherClass):
    def __init__(self, *args, **kwargs):
        super(self.__class__, self).__init__(*args, **kwargs)