Using super with a class method

dza picture dza · Nov 30, 2009 · Viewed 51k times · Source

I'm trying to learn the super() function in Python.

I thought I had a grasp of it until I came over this example (2.6) and found myself stuck.

http://www.cafepy.com/article/python_attributes_and_methods/python_attributes_and_methods.html#super-with-classmethod-example

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "test.py", line 9, in do_something
    do_something = classmethod(do_something)
TypeError: unbound method do_something() must be called with B instance as first argument (got nothing instead)
>>>

It wasn't what I expected when I read this line right before the example:

If we're using a class method, we don't have an instance to call super with. Fortunately for us, super works even with a type as the second argument. --- The type can be passed directly to super as shown below.

Which is exactly what Python tells me is not possible by saying that do_something() should be called with an instance of B.

Answer

unutbu picture unutbu · Nov 30, 2009

Sometimes texts have to be read more for the flavor of the idea rather than for the details. This is one of those cases.

In the linked page, Examples 2.5, 2.6 and 2.7 should all use one method, do_your_stuff. (That is, do_something should be changed to do_your_stuff.)

In addition, as Ned Deily pointed out, A.do_your_stuff has to be a class method.

class A(object):
    @classmethod
    def do_your_stuff(cls):
        print 'This is A'

class B(A):
    @classmethod
    def do_your_stuff(cls):
        super(B, cls).do_your_stuff()

B.do_your_stuff()

super(B, cls).do_your_stuff returns a bound method (see footnote 2). Since cls was passed as the second argument to super(), it is cls that gets bound to the returned method. In other words, cls gets passed as the first argument to the method do_your_stuff() of class A.

To reiterate: super(B, cls).do_your_stuff() causes A's do_your_stuff method to be called with cls passed as the first argument. In order for that to work, A's do_your_stuff has to be a class method. The linked page doesn't mention that, but that is definitively the case.

PS. do_something = classmethod(do_something) is the old way of making a classmethod. The new(er) way is to use the @classmethod decorator.


Note that super(B, cls) can not be replaced by super(cls, cls). Doing so could lead to infinite loops. For example,

class A(object):
    @classmethod
    def do_your_stuff(cls):
        print('This is A')

class B(A):
    @classmethod
    def do_your_stuff(cls):
        print('This is B')
        # super(B, cls).do_your_stuff()  # CORRECT
        super(cls, cls).do_your_stuff()  # WRONG

class C(B):
    @classmethod
    def do_your_stuff(cls):
        print('This is C')
        # super(C, cls).do_your_stuff()  # CORRECT
        super(cls, cls).do_your_stuff()  # WRONG

C.do_your_stuff()

will raise RuntimeError: maximum recursion depth exceeded while calling a Python object.

If cls is C, then super(cls, cls) searches C.mro() for the class that comes after C.

In [161]: C.mro()
Out[161]: [__main__.C, __main__.B, __main__.A, object]

Since that class is B, when cls is C, super(cls, cls).do_your_stuff() always calls B.do_your_stuff. Since super(cls, cls).do_your_stuff() is called inside B.do_your_stuff, you end up calling B.do_your_stuff in an infinite loop.

In Python3, the 0-argument form of super was added so super(B, cls) could be replaced by super(), and Python3 will figure out from context that super() in the definition of class B should be equivalent to super(B, cls).

But in no circumstance is super(cls, cls) (or for similar reasons, super(type(self), self)) ever correct.