Can I use Python 3 super() in Python 2.5.6?

Question

Can I use clean Python 3 super() syntax in Python 2.5.6?
Maybe with some kind of __future__ import?

Fortalice

Answer 1

You cannot use a bare super() call that contains no type/class. Nor can you implement a replacement for it that will work. Python 3.x contains special support to enable bare super() calls (it places a __class__ cell variable in all functions defined within a class - see PEP 3135

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Update

As of Python 2.6+, bare super() calls can be used via the future Python package. See posita's answer for an explanation.

Lyudmila

Thanks. I was confused because an earlier version of this PEP said you'd import it with from __future__ import new_super which doesn't work. – Fortalice

Answer 2

I realize this question is old, and the selected answer may have been correct at the time, but it's no longer complete. You still can't use super() in 2.5.6, but python-future provides a back-ported implementation for 2.6+:

Install python-future with:

% pip install future

The following shows the redefinition of super under builtins:

% python
...
>>> import sys
>>> sys.version_info[:3]
(2, 7, 9)
>>>
>>> super
<type 'super'>
>>>
>>> from builtins import *
>>> super
<function newsuper at 0x000000010b4832e0>
>>> super.__module__
'future.builtins.newsuper'

It can be used as follows:

from builtins import super

class Foo(object):
    def f(self):
        print('foo')

class Bar(Foo):
    def f(self):
        super().f() # <- whoomp, there it is
        print('bar')

b = Bar()
b.f()

which outputs

foo
bar

If you use pylint, you can disable legacy warnings with the comment:

# pylint: disable=missing-super-argument

Chronicle

Answer 3

No you cannot. But you can use Python 2's super() in Python 3.

Spriggs

Answer 4

Note This is a terrible "solution", I post it only to make sure you don't do this at home!
I repeat: do not do this

One may think about using this mixin

class Super(object):
    def super(self):
        return super(self.__class__, self)

to obtain a self.super():

class A(object, Super):
    def __init__(self):
        print "A"

class B(A):
    def __init__(self):
        print "B"
        self.super().__init__()

yielding:

 >>> a = A()
 A
 >>> b = B()
 B
 A

But beware: This self.super() is not equivalent to super(B, self) - if A also called self.super().__init__(), the construction of a B would cause the A constructor to call itself indefinitely, since self.__class__ will remain B. This is due to the lack of the __class__ mentioned in the accepted answer. You can possibly work around this issue with a hidden state machine or a sophisticated metaclass that e.g. checks the actual class's position in self.__class__.mro(), but is it really worth it? Probably not...

Uno