Why do I have to specify my own class when using super(), and is there a way to get around it?

Question

When using Python\'s super() to do method chaining, you have to explicitly specify your own class, for example:

class MyDecorator(Decorator):
           


        

Answer 1

In Python 3.0, you can use super() which is equivalent to super(ThisClass, self).

Documentation here. Code sample from the documentation:

class C(B):
    def method(self, arg):
        super().method(arg)    
        # This does the same thing as: super(C, self).method(arg)

Answer 2

This answer is wrong, try:

def _super(cls):
    setattr(cls, '_super', lambda self: super(cls, self))
    return cls

class A(object):
    def f(self):
        print 'A.f'

@_super
class B(A):
    def f(self):
        self._super().f()

@_super
class C(B):
    def f(self):
        self._super().f()

C().f() # maximum recursion error

---

In Python 2 there is a way using decorator:

def _super(cls):
    setattr(cls, '_super', lambda self: super(cls, self))
    return cls

class A(object):
    def f(self):
        print 'A.f'

@_super
class B(A):
    def f(self):
        self._super().f()

B().f() # >>> A.f

Answer 3

you can also avoid writing a concrete class name in older versions of python by using

def __init__(self):
    super(self.__class__, self)
    ...

Answer 4

The BDFL agrees. See PEP 3135 - New Super for Python 3.0 (and Pep 367 - New Super for Python 2.6).