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:

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Answer 1

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 2

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. :-)

Answer 3

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.