New submission from William Schwartz <wkschwa...@gmail.com>: 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.
---------- assignee: docs@python components: Documentation, Tests messages: 164606 nosy: William.Schwartz, docs@python priority: normal severity: normal status: open title: Misleading documentation for __prepare__ type: behavior versions: Python 3.3 _______________________________________ Python tracker <rep...@bugs.python.org> <http://bugs.python.org/issue15243> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: http://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
