Provide __classcell__ example for Python 3.6 metaclass

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别那么骄傲
别那么骄傲 2020-12-03 17:37

Per the 3.6.0 docs:

CPython implementation detail: In CPython 3.6 and later, the __class__ cell is passed to the metac

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  • 2020-12-03 18:10

    The warning is raised if you use super that relies on __class__ being available or reference __class__ inside the class body.

    What the text essentially says is that, this is needed if you define a custom meta-class and tamper with the namespace you get before passing it up to type.__new__. You'll need to be careful and always make sure you pass __classcell__ to type.__new__ in your metaclass.__new__.

    That is, if you create a new fancy namespace to pass up, always check if __classcell__ is defined in the original namespace created and add it:

    class MyMeta(type):
        def __new__(cls, name, bases, namespace):
            my_fancy_new_namespace = {....}  
            if '__classcell__' in namespace:
                 my_fancy_new_namespace['__classcell__'] = namespace['__classcell__']
            return super().__new__(cls, name, bases, my_fancy_new_namespace)
    

    The file you linked in the comment is actually the first of many attempted patches, issue23722_classcell_reference_validation_v2.diff is the final patch that made it in, from Issue 23722.

    An example of doing this correctly can be seen in a pull request made to Django that uses this to fix an issue that was introduced in Python 3.6:

    new_attrs = {'__module__': module}
    classcell = attrs.pop('__classcell__', None)
    if classcell is not None:
        new_attrs['__classcell__'] = classcell
    new_class = super_new(cls, name, bases, new_attrs)
    

    The __classcell__ is simply added to the new namespace before being passed to type.__new__.

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