On 10/27/2018 2:53 PM, Joy Diamond wrote:
Chris,
Yes, the following works:
"""
(Note that any object `x` is always considered to be an instance of
`type(x)`, and this cannot be overridden.)
"""
Open a doc issue on bugs.python.org with the problem, motivation, and
proposed solution and it should get considered.
NOTE: I fixed your sentence of `x.__class__` to `type(x)` since it is
not always true that `x.__class__ == type(x)`.
For example the actual code reference above:
https://github.com/python/cpython/blob/master/Objects/abstract.c#L2397-L2405
Says "if (Py_TYPE(inst) == (PyTypeObject *)cls)" in the actual C Python
implementation:
So it using `type(x)` not `x.__class__`
Thanks,
Joy Diamond.
====
ADDENDUM:
Here is an example where `type(x) is not x.__class__` (there are other
not as perverse examples where you want `.__class__` to differ from `type`)
NOTE: The final assert will fail, showing that `type(x) is not x.__class__`
#
# This really perverse code demonstrates that `t.__class__` is *NOT*
# always the same as type(t).
#
# The code shows an metaclass that when you derive objects from its
classes, creates a `7`
# instead of a derived class.
#
def not_really_a_metaclass__make_a_7(name, bases, member):
return 7
@property
def not_really_a_class__make_a_7(t):
return not_really_a_metaclass__make_a_7
class Metaclass__Make_A_7(type):
__class__ = not_really_a_class__make_a_7
Make_A_7 = Metaclass__Make_A_7('Make_A_7', ((object,)), {})
#
# Now then:
#
# `Make_A_7` is a class that when inherited from creates a '7' instead
# of a class ... :(
#
# Works for python 2 & pypy (not yet fixed to work for python 3)
#
class Seven(Make_A_7): # Calls `Make_A_7.__class__('Seven,
(('Make_A_7,)), {})` which returns `7`
pass
print('Seven is: %s' % Seven)
assert isinstance(Make_A_7, Metaclass__Make_A_7)
assert Make_A_7.__class__ == Metaclass__Make_A_7 # This will *FAIL*
due to the perverse definition of `.__class__`
On Sat, Oct 27, 2018 at 2:25 PM Chris Angelico
<ros...@gmail.com
<mailto:ros...@gmail.com>> wrote:
On Sun, Oct 28, 2018 at 5:03 AM Joy Diamond
<python....@gmail.com
<mailto:python....@gmail.com>> wrote:
>
> Greetings,
>
> This is a request to fix the documentation for __instancecheck__.
>
> Please add the following (please rewrite better than I can -- I
am not good at explaining concepts in short sentences):
>
> NOTE: As an optimization, isinstance(object, classinfo) does NOT
call classinfo.__instancecheck__(instance) when type(object) ==
classinfo.
>
> Consider the following program:
>
> class M(type):
> def __instancecheck__(m, t):
> print('instancecheck(%s, %s)' % (m, t))
> return False # LIE!
>
> Test = M('Test', ((object,)), {})
>
> something = Test()
>
> print('Does *NOT* call __instancecheck__:')
> print('isinstance(something, Test): %s' % isinstance(something,
Test))
Here's the passage in question, for reference:
"""
The following methods are used to override the default behavior of the
isinstance() and issubclass() built-in functions.
In particular, the metaclass abc.ABCMeta implements these methods in
order to allow the addition of Abstract Base Classes (ABCs) as
“virtual base classes” to any class or type (including built-in
types), including other ABCs.
"""
https://docs.python.org/3/reference/datamodel.html#customizing-instance-and-subclass-checks
Since it uses the word "override", I agree that it's not entirely
correct. The implication of "override" is that you can completely
replace the normal behaviour. In this case, you can change the
behaviour of subclass testing (for instance, you can "disown" a
subclass by denying that instances of it are instances of yourself),
and of course, you can claim an object as an instance of a class it
didn't directly inherit from (the way ABCs work), but you cannot fib
about direct instances. I think the behaviour is close enough to
accurate that it doesn't need major rewording; how about adding this
parenthesis:
"""
(Note that any object `x` is always considered to be an instance of
`x.__class__`, and this cannot be overridden.)
"""
Would that work?
ChrisA
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