python - How to create the int 1 at two different memory locations?

Question

I want to show someone how using is instead of == to compare integers can fail. I thought this would work, but it didn't:

>>> import copy
>>> x = 1
>>> y = copy.deepcopy(x)
>>> x is y
True

I can do this easily for bigger integers:

>>> x = 500
>>> y = 500
>>> x is y
False

How can I demonstrate the same thing with smaller integers which might typically be used for enum-like purposes in python?

Answer 1

The following example fails in both Python 2 and 3:

>>> n=12345
>>> ((n**8)+1) % (n**4) is 1
False
>>> ((n**8)+1) % (n**4) == 1
True

The reasons are slightly different. Python 2 uses the int type for small integers and the long type for arbitrary precision values. Only the int type is interned so the example fails when a 1L is returned.

Python 3 only uses the arbitrary precision type (and renamed it to int). The example fails because the remainder calculation internally computes a value of 1 and returns it. The interning check is only done when objects are created and the object was created at the start of the calculation before it had the value 1.