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

问题

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

class MyDecorator(Decorator):
    def decorate(self):
        super(MyDecorator, self).decorate()

I have to specify the name of my class MyDecorator as an argument to super(). This is not DRY. When I rename my class now I will have to rename it twice. Why is this implemented this way? And is there a way to weasel out of having to write the name of the class twice(or more)?

回答1:

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

回答2:

Your wishes come true:

Just use python 3.0. In it you just use super() and it does 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)

回答3:

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

回答4:

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

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

来源：https://stackoverflow.com/questions/466611/why-do-i-have-to-specify-my-own-class-when-using-super-and-is-there-a-way-to

标签

python

multiple-inheritance