Python “is” statement: unexpected behavior when comparing unbound method

Asked Aug 18 '15 at 06:19

Active Aug 18 '15 at 06:19

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In Python 2.7.9, when I assign a unbound method to a new attribute, and compare them by is statement, the result is False:

In [1]: class A(object):
   ...:     def a(self):
   ...:         pass
   ...:

In [2]: A._a = A.a

In [3]: print A.a, A._a
<unbound method A.a> <unbound method A.a>

In [4]: print id(A.a), id(A._a)
4499595904 4499595904

In [5]: A.a is A._a
Out[5]: False

This is very counter-intuitive and I could not find any reference or documentation to explain this behavior. What's more, when I test the same code in Python 3.4.2, the result turned to be True. I'm guessing that it's a bug in Python 2.7 but got fixed in Python 3, can anyone help me find the actual reason why this happen?

asked Aug 18 '15 at 06:19

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mb it is because a is a method and _a is link. – Dmitry.Samborskyi Aug 18 '15 at 06:24
This question is gonna be good.. – Cyphase Aug 18 '15 at 06:24
1

This is the answer I think: https://stackoverflow.com/questions/13348031/python-bound-and-unbound-method-object – Cyphase Aug 18 '15 at 06:27
As for why the IDs are the same, see for example [this question](http://stackoverflow.com/questions/3877230/why-does-id-id-and-id-id-in-cpython). – poke Aug 18 '15 at 06:29

Python “is” statement: unexpected behavior when comparing unbound method

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