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Date: Tue, 19 Apr 2011 17:45:42 -0700
Subject: Re: My stupidity / strange inconsistency overriding class methods
From: Chris Rebert <clp2@rebertia.com>
Cc: python-list@python.org
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On Tue, Apr 19, 2011 at 4:52 PM, andrew cooke <andrew@acooke.org> wrote:
> Hi,
>
> I've been staring at this problem, in various forms, all day. =C2=A0Am I =
missing something obvious, or is there some strange hardwiring of isinstanc=
e? =C2=A0This is with Python 3.2.
>
> =C2=A0 =C2=A0 =C2=A0 =C2=A0class A(metaclass=3DABCMeta):
> =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0@classmethod
> =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0def __instancecheck__(cls, insta=
nce): return False
> =C2=A0 =C2=A0 =C2=A0 =C2=A0# no override
> =C2=A0 =C2=A0 =C2=A0 =C2=A0assert isinstance(A(), A)
> =C2=A0 =C2=A0 =C2=A0 =C2=A0assert A.__class__.__instancecheck__(A, A())

[You've already figured out the issue, but since I spent a while
composing this, and for the benefit for the archives, I'll post
anyway.]

Makes sense after a little thought.
http://docs.python.org/reference/datamodel.html#customizing-instance-and-su=
bclass-checks
"Note that [ __instancecheck__() is ] looked up on the type
(metaclass) of a class. [It] cannot be defined as [a classmethod] in
the actual class. This is consistent with the lookup of special
methods that are called on instances, only in this case the instance
is itself a class."

Recall from http://docs.python.org/reference/datamodel.html#special-method-=
lookup-for-new-style-classes
that lookup of __special__ methods never consults instance
dictionaries, instead skipping directly to the type's namespace; as
the quote says, in this case, the instance (of ABCMeta) is itself a
class/type (namely A). Your two assert statements are therefore almost
precisely equivalent in this case; and since the latter involves
A.__class__ (a.k.a. ABCMeta) rather than A itself, it's understandable
that that A's namespace is not consulted.

> =C2=A0 =C2=A0 =C2=A0 =C2=A0class B(type):
> =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0def foo(self): return 42
> =C2=A0 =C2=A0 =C2=A0 =C2=A0class C(metaclass=3DB):
> =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0@classmethod
> =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0 =C2=A0def foo(cls): return 7
> =C2=A0 =C2=A0 =C2=A0 =C2=A0# override
> =C2=A0 =C2=A0 =C2=A0 =C2=A0assert C().__class__.foo() =3D=3D 7

More simply:  assert C.foo() =3D=3D 7

"foo" is not a __special__ method name; therefore we look in the
instance dictionary of the receiver (i.e. C) before consulting the
receiver's type (i.e. B). Our check in the instance dictionary is
successful (we find C.foo), and therefore we don't even bother looking
at C's type (i.e. B, where we would find B.foo).

> It seems to me that the above two cases are inconsistent. =C2=A0ABCMeta d=
eclares __instancecheck__ just like B declares foo. =C2=A0Yet C can overrid=
e foo, but A is unable to override the instance check.

The difference is in the __special__-ness of the method names in question.

Cheers,
Chris
--
http://blog.rebertia.com