Python extending with - using super() Python 3 vs Python 2

Question

Originally I wanted to ask this question, but then I found it was already thought of before...

Googling around I found this example of extending configparser. The following works with Python 3:

$ python3
Python 3.2.3rc2 (default, Mar 21 2012, 06:59:51) 
[GCC 4.6.3] on linux2
>>> from configparser import  SafeConfigParser
>>> class AmritaConfigParser(SafeConfigParser):
...     def __init__(self):
...         super().__init__()
... 
>>> cfg = AmritaConfigParser()

But not with Python 2:

>>> class AmritaConfigParser(SafeConfigParser):
...       def __init__(self):
...           super(SafeConfigParser).init()
... 
>>> cfg = AmritaConfigParser()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 3, in __init__
TypeError: must be type, not classob

Then I read a little bit on Python New Class vs. Old Class styles (e.g. here. And now I am wondering, I can do:

class MyConfigParser(ConfigParser.ConfigParser):
      def Write(self, fp):
          """override the module's original write funcition"""
          ....
      def MyWrite(self, fp):
          """Define new function and inherit all others"""

But, shouldn't I call init? Is this in Python 2 the equivalent:

 class AmritaConfigParser(ConfigParser.SafeConfigParser):
    #def __init__(self):
    #    super().__init__() # Python3 syntax, or rather, new style class syntax ...
    #
    # is this the equivalent of the above ? 
    def __init__(self):
        ConfigParser.SafeConfigParser.__init__(self)

Answer 1

• super() (without arguments) was introduced in Python 3 (along with __class__):

  super() -> same as super(__class__, self)
  

  so that would be the Python 2 equivalent for new-style classes:

  super(CurrentClass, self)
  

• for old-style classes you can always use:

   class Classname(OldStyleParent):
      def __init__(self, *args, **kwargs):
          OldStyleParent.__init__(self, *args, **kwargs)