with_metaclass: check existence of `__prepare__` attribute before calling it

[behnam]

https://github.com/prettier/plugin-python/blob/04dc40a40501a13eaec5fcb4d5c99829a31d3ffc/vendor/python/six.py#L831

The def with_metaclass() implementation assumes that all meta classes have __prepare__ defined.

The Python Reference says (https://docs.python.org/3/reference/datamodel.html#preparing-the-class-namespace):

If the metaclass has a __prepare__ attribute, it is called as [...].

If the metaclass has no __prepare__ attribute, then the class namespace is initialised as an empty ordered mapping.

Therefore, I think six should check existence of __prepare__ attribute before calling it, as __prepare__ is not defined on type in PY2.

[benjaminp]

Why is that a problem, though? Defining a __prepare__ method on Python 2 is useless but harmless, right?

[behnam]

Looks like Cython makes calls to __prepare__ even in Python 2. So, not 100% a bug here, but would be great to address it so application (like qcore here) don't have to special-case for it.

[benjaminp]

Is there a Cython bug filed for that behavior?

[behnam]

Here's the original report on Cython: cython/cython#1936

The issue was introduced in #178, already reported in #210, and an Open PR is out in #211. Unfortunately both the issue and PR have been open for 11 months, without much movement.

@benjaminp, can we please either merge in #211, or go with a quick (and simpler) fix, which would be calling hasattr() before calling __prepare__()?

[aaronchall]

This is a problem because Python accepts any callable with the right signature as a metaclass, and this assumes that you are subclassing type (or something with a __prepare__ method) - so this behavior could be considered buggy for that reason.

What I don't understand is why you're defining a metaclass in the decorator - why don't you just return

meta('temporary_class', bases, {})