Confusion about Python operator 'is'

Question

Definition of 'is' operator in python:

is operator checks whether both the operands refer to the same object or not

Then how come when the id of a and list1[0] don't match, the 2nd condition is True?

Does this answer your question? "is" operator behaves unexpectedly with integers — rassar
– rassar, Commented Mar 28, 2020 at 23:13
No I know about that case. Python keeps an array of integer objects for all integers between -5 and 256. When you create an integer in that range, you get back a reference to the already existing object. Anything outside of that creates a new reference. — Shafin M
– Shafin M, Commented Mar 28, 2020 at 23:16

rdas · Accepted Answer · 2020-03-28 23:17:16Z

4

What you're doing when you're doing id(a) is id(list1[0]) is comparing the values returned by the id() function to check if they point to the same object or not. Those values are different objects - EVEN IF THEY ARE THE SAME VALUE

Check this:

a = 2
ll = [2]

print(a is ll[0])
print(id(a), id(ll[0]))
print(id(a) is id(ll[0]))

Which gives:

True
140707131548528 140707131548528
False

Now why is the first result True? Because of interning - all ints between -5 & 256 are pre-created objects that are reused. So every 2 in python is actually the same object. But every 140707131548528 is different

answered Mar 28, 2020 at 23:17

rdas

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1 Comment

Shafin M Over a year ago

Thanks for the explanation. I'm still finding this pretty confusing but I'll get the hang of it.

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