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

Question

I had a bug where I was relying on methods being equal to each other when using is. It turns out that's not the case:

>>> class What:
...     def meth(self):
...         pass

>>> What.meth is What.meth
True
>>> inst = What()
>>> inst.meth is inst.meth
False

Why is that the case? It works for regular functions:

>>> def func(): pass
>>> func is func
True

Answer 1

Method objects are created each time you access them. Functions act as descriptors, returning a method object when their .__get__ method is called:

>>> What.__dict__['meth']
<function What.meth at 0x10a6f9c80>
>>> What.__dict__['meth'].__get__(What(), What)
<bound method What.meth of <__main__.What object at 0x10a6f7b10>>

If you're on Python 3.8 or later, you can use == equality testing instead. On Python 3.8 and later, two methods are equal if their .__self__ and .__func__ attributes are identical objects (so if they wrap the same function, and are bound to the same instance, both tested with is).

Before 3.8, method == behaviour is inconsistent based on how the method was implemented - Python methods and one of the two C method types compare __self__ for equality instead of identity, while the other C method type compares __self__ by identity. See Python issue 1617161.

If you need to test that the methods represent the same underlying function, test their __func__ attributes:

>>> What.meth == What.meth     # functions (or unbound methods in Python 2)
True
>>> What().meth == What.meth   # bound method and function
False
>>> What().meth == What().meth # bound methods with *different* instances
False
>>> What().meth.__func__ == What().meth.__func__ # functions
True