Comparing two variables with 'is' operator which are declared in one line in Python [duplicate]

Question

According to the Documentation:

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. :-)

So the following behaviors are normal.

>>> a = 256
>>> b = 256
>>> a is b
True
>>> c = 257
>>> d = 257
>>> c is d
False

But when i declare two variables like these, i am getting True-

>>> e = 258; f=258;
>>> e is f
True

I have checked the identity of the objects referenced by e and f-

>>> id(e)
43054020
>>> id(f)
43054020

They are same.

My question is what is happening when we are declaring e and f by separating with semicolons? Why are they referencing to the same object (though the values are out of the range of Python's array of integer objects) ?

It would be better, if you please explain it like you are explaining it to a beginner.

Answer 1

This is not an unexpected behavior, according to Python Data model it's an implementation detail:

Types affect almost all aspects of object behavior. Even the importance of object identity is affected in some sense: for immutable types, operations that compute new values may actually return a reference to any existing object with the same type and value, while for mutable objects this is not allowed. E.g., after a = 1; b = 1, a and b may or may not refer to the same object with the value one, depending on the implementation, but after c = []; d = [], c and d are guaranteed to refer to two different, unique, newly created empty lists. (Note that c = d = [] assigns the same object to both c and d.)