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Re: Calling super() in __init__ of a metaclass

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  Re: Calling super() in __init__ of a metaclass Eli Bendersky <eliben@gmail.com> - 2011-08-06 11:18 +0300

#10966 — Re: Calling super() in __init__ of a metaclass

On Sat, Aug 6, 2011 at 11:04, Chris Rebert <clp2@rebertia.com> wrote:
> On Sat, Aug 6, 2011 at 12:34 AM, Eli Bendersky <eliben@gmail.com> wrote:
>> Consider this standard metaclass definition:
>>
>> class MyMetaclass(type):
>>    def __init__(cls, name, bases, dct):
>>        super(MyMetaclass, cls).__init__(name, bases, dct)
>>        # do meta-stuff
>>
>> class Foo(object):
>>    __metaclass__ = MyMetaclass
>>
>> The call "super(MyMetaclass, cls)" should returns the parent of
>> MyMetaclass here. But the 'cls' passed into this __init__ is *not*
>> MyMetaclass, but rather the created class - i.e. Foo.
>
> ...which is an instance of the first argument to super(), which is how
> super is typically invoked. Remember: a class is an instance of its
> metaclass.
>
> Perhaps if you rename `cls` to `self` and then re-read your snippet,
> you'll have a flash of sudden understanding. Calling the parameter
> `cls` is merely convention.
>
>> So how does
>> "super" get to the parent of MyMetaclass using this information? The
>> documentation of "super" says:
>>
>>    If the second argument is a type, issubclass(type2, type) must be
>> true (this is useful for classmethods).
>>
>> Yes, 'cls' is a type (it's "class Foo"), but no, it's not a subclass
>> of MyMetaclass, so this doesn't help.
>
> The typical form of super(), mentioned earlier in the documentation,
> is being used here:
>
>    super(type, obj) -> bound super object; requires isinstance(obj, type)
>
> `type` here is the metaclass, `obj` here is the class. By definition,
> a class is an *instance* of its metaclass, so the precondition is
> satisfied.

Thanks, Chris. This clarifies things.

Eli

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