Section 3.3.3.2. "Preparing the class namespace" of the documentation (http://docs.python.org/dev/reference/datamodel.html#preparing-the-class-namespace) states that "If the metaclass has a __prepare__ attribute, it is called as ``namespace = metaclass.__prepare__(name, bases, **kwds)``...." This isn't quite true. By just defining a ``__prepare__`` method in a metaclass, the interpreter calls it as it would a static method -- there is no implicit first argument referring to ``metaclass`` as the documentation implies. The documentation should be amended to say that users can decorate ``__prepare__`` as a class method to get ``metaclass`` passed in as the implicit first argument.

Attached a unittest script to demonstrate that __prepare__ is implicitly a staticmethod.

__prepare__ is not implicitly a staticmethod, and it is called exactly as documented (also in types.prepare_class). There is no implicit first argument because the method is called on the (meta)class object.

Daniel, Good point. However it would still be useful for documentation to point out that __prepare__ can be passed the metaclass as the implicit first argument by being decorated by classmethod.

I'll post a small patch when I get a chance to add a sentence saying as much in the documentation and reorganize the test cases in Lib/test/test_metaclass.py to make it part of the narrative of the doctests.

Actually the docs contained a similar sentence ("If the metaclass has a :meth:`__prepare__` attribute (usually implemented as a class or static method), ..."), but it was removed in befd56673c80 when Nick rewrote this section.

Does the documentation need amending, yes or no?

I think the current documentation is correct and doesn't need changes. There is also already an example of a working __prepare__ method.

The documentation should mention that the __prepare__ method must be implemented as a classmethod (see msg164697 for the reference.)

The example at https://docs.python.org/dev/reference/datamodel.html#metaclass-example needs to be linked from the "Preparing the class namespace" section.

Submitting patch according to this (http://bugs.python.org/issue15243#msg268356) message. Thanks!

Is this still open, does the patch still need a pull request?

New changeset 7de617455ed788e6730c40cf854c4b72b0432194 by Berker Peksag (alclarks) in branch 'master':
bpo-15243: Document __prepare__ as classmethod (GH-17124)
https://github.com/python/cpython/commit/7de617455ed788e6730c40cf854c4b72b0432194

New changeset 321491a536c378227f9d574703f7c06f89c67dcf by Berker Peksag (Miss Islington (bot)) in branch '3.8':
bpo-15243: Document __prepare__ as classmethod (GH-17124)
https://github.com/python/cpython/commit/321491a536c378227f9d574703f7c06f89c67dcf

New changeset 548685e364af3d366d3aef924b2077ee6f8d13dc by Berker Peksag (Miss Islington (bot)) in branch '3.7':
bpo-15243: Document __prepare__ as classmethod (GH-17124)
https://github.com/python/cpython/commit/548685e364af3d366d3aef924b2077ee6f8d13dc