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.