Python's super() considered super!

[Ian Kelly]

On Thu, May 26, 2011 at 10:31 AM, Raymond Hettinger <python at rcn.com> wrote:
> I just posted a tutorial and how-to guide for making effective use of
> super().

I posted this already on the HackerNews thread but it seems to have
largely gone unnoticed, so I'm reposting it here.

It seems to me that the example of combining built-in dictionary
classes is naively optimistic. For starters, OrderedDict, as it
happens, does not use super! It calls the dict super-class methods
directly. Since dict luckily happens to be the next class in the MRO,
this doesn't really matter for the purpose of this example, but I can
envision a scenario where some plucky programmer inherits from both
OrderedCounter and some other dict subclass, and the result doesn't
work because OrderedDict does the wrong thing.

And OrderedDict isn't the only one. Maybe for some reason I would like
to have an OrderedCounter where all the counts default to 42. So I do
this:

  class DefaultOrderedCounter(defaultdict, OrderedCounter):
      pass
  doc = DefaultOrderedCounter(lambda: 42)
  doc.update('abracadabra')

Which results in:

  Traceback (most recent call last):
    File "<stdin>", line 1, in <module>
    File "c:\python32\lib\collections.py", line 507, in update
      _count_elements(self, iterable)
    File "c:\python32\lib\collections.py", line 63, in __setitem__
      self.__map[key] = link = Link()
  AttributeError: 'DefaultOrderedCounter' object has no attribute
'_OrderedDict__map'

Whoops! Apparently defaultdict doesn't use super either. Of course a
better way to do this would be to subclass OrderedCounter and just
override the __missing__ method by hand, but that's not the point.

The article goes into "How to Incorporate a Non-cooperative Class",
which basically says "wrap it up in a proxy class". But that's not
really going to work here, since the result would be two separate
dicts, with the defaultdictwrapper methods operating on one dict, and
the other methods operating on the other.