How to create the int 1 at two different memory locations?
cpython, python
Solution
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.
Problem
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?