Why is a method not identical to itself?

Question

The Python documentation about the is operator says:

The operators is and is not test for
object identity: x is y is true if and
only if x and y are the same object.
x is not y yields the inverse truth
value.

Let’s try that:

>>> def m():
...   pass
... 
>>> m is m
True

The Python documentation also says:

Due to automatic garbage-collection,
free lists, and the dynamic nature of
descriptors, you may notice seemingly
unusual behaviour in certain uses of
the is operator, like those involving
comparisons between instance methods,
or constants. Check their
documentation for more info.

>>> class C:
...   def m():
...     pass
... 
>>> C.m is C.m
False

I searched for more explanations, but I was not able to find any.

Why is C.m is C.m false?

I am using Python 2.x. As noted in the answers below, in Python 3.x C.m is C.m is true.

Asked By: Lesmana

||

Source

Answer 1

When you ask for an instance attribute which is a function, you get a bound method: a callable object which wraps the function defined in the class and passes the instance as the first argument. In Python 2.x, when you ask for a class attribute which is a function, you get a similar proxy object called an unbound method:

>>> class A:
...     def m():
...         return None
... 
>>> A.m
<unbound method A.m>

This special object is created when you ask for it, and not apparently cached anywhere. That means that when you do

>>> A.m is A.m
False

you are creating two distinct unbound method objects and testing them for identity.

Notice that these work fine:

>>> x = A.m
>>> x is x
True

and

>>> A.m.im_func is A.m.im_func
True

(im_func is the original function which the unbound method object is wrapping.)

In Python 3.x, incidentally, C.m is C.m is True, because the (somewhat pointless) unbound method proxy objects were removed entirely and you just get the original function which you defined.

This is just one example of the very dynamic nature of attribute lookup in Python: when you ask for an attribute of an object, it is possible to run arbitrary code to calculate the value of that attribute. Here’s another example where your test fails in which it is much clearer why:

>>> class ChangingAttribute(object):
...     @property
...     def n(self):
...         self._n += 1
...         return self._n
... 
...     def __init__(self):
...         self._n = 0
... 
>>> foo = ChangingAttribute()
>>> foo.n
1
>>> foo.n
2
>>> foo.n
3
>>> foo.n is foo.n
False
>>> foo.n
6

Answered By: Katriel

Answer 2

Because C.m() is not a static method of the class C:

Try it like this:

class C:
    @staticmethod
    def m():
        pass

print C.m is C.m
# True

c = C()
print c.m is C.m
# True

Because static methods are like class variables, we only want one reference for them so that if we change their bound value this change should be automatic in all the class and instance of this class.

On the other hand, in your example, C.m is not a static method so Python makes the assumption that it should be treated like a non-static method, so whenever you call C.m, it will return a new instance:

class C:
   def m():
      pass

a = C.m
b = C.m

print id(a), id(b)
# 43811616, 43355984
print a is b
# False

N.B: static methods are not like class methods!

Answered By: mouad

Answer 3

I assume you are using Python 2? In Python 3, C.m is C.m (but C().m is C().m is still false). If you enter just C.m at the REPL, I bet you see something like <UnboundMethod... >. An UnboundMethod wrapper does very little, except checking isinstance(self, cls). (Seems pretty pointless to create a wrapper for this? It is, so it was dropped in Python 3 – C.m is just a function). A fresh wrapper instance is created on-demand whenever the method is accessed – C.m creates one, another C.m creates another one. Since they’re different instances, C.m is not C.m.

Closely related are the bound methods, which allow you to do f = obj.method; f(*args) but also cause instance.method is not instance.method. Upon instanciation, all functions defined in the class (read: all methods, except of course monkeypatched ones) become properties of the instance. When you access them, you instead get a fresh instance of a wrapper (the bound method) around the plain function. This wrapper remembers the instance (self) and when called with (arg1, arg2, ..., argN) just hands these on to the function – with self added as first argument. You usually don’t notice because you call the method right away – but this is what allows passing self implicitly without resorting to language-level trickery.

See the history of Python for more details and, well, history.

Answered By: user395760

Why is a method not identical to itself?

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