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.

Answer 2

You can do:

>>> 0 - 6 is -6
False
>>> 0 - 6 == -6
True

It also works for bigger numbers:

>>> 1000 + 1 is 1001
False
>>> 1000 + 1 == 1001
True

It depends on what you want to demonstrate but the above highlights the difference in functionality between is and ==.

Answer 3

What you observe is expected:

>>> x=256
>>> y=256
>>> x is y
True
>>> x=257
>>> y=257
>>> x is y
False
>>> x=-5
>>> y=-5
>>> x is y
True
>>> x=-6
>>> y=-6
>>> x is y
False

Quoting from Plain Integer Objects:

The current implementation keeps an array of integer objects for all integers between -5 and 256, when you create an int in that range you actually just get back a reference to the existing object. So it should be possible to change the value of 1. I suspect the behaviour of Python in this case is undefined. :-)