I want to know how Python knows (if it knows) that a value-type object is already stored in its memory (and also knows where it is).

For this code, when assigning the value 1 for b, how does it know that the value 1 is already in its memory and stores its reference in b?

>>> a = 1
>>> b = 1
>>> a is b
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    Use print(hex(id(b))) to check memory address for b – Yusufsn Apr 19 at 3:01
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    >>> hex(id(b))'0x7ffe705ee350' >>> hex(id(a)) '0x7ffe705ee350' – Just A Lone Apr 19 at 3:03
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    If two variables refer to the same value between -5 and 256 (as opposed to use) then by definition there is only one object. – Yusufsn Apr 19 at 3:04
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    @Yusufsn No. For bigger integers (>256) it's not true. – asn-0184 Apr 19 at 3:09
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    As I said, only values between -5 and 256 – Yusufsn Apr 19 at 3:11

Python (CPython precisely) uses shared small integers to help quick access. Integers range from [-5, 256] already exists in memory, so if you check the address, they are the same. However, for larger integers, it's not true.

a = 100000
b = 100000
a is b # False

Wait, what? If you check the address of the numbers, you'll find something interesting:

a = 1
b = 1
id(a) # 4463034512
id(b) # 4463034512

a = 257
b = 257
id(a) # 4642585200
id(b) # 4642585712

It's called integer cache. You can read more about the integer cache here.

Thanks comments from @KlausD and @user2357112 mentioning, direct access on small integers will be using integer cache, while if you do calculations, though they might equals to a number in range [-5, 256], it's not a cached integer. e.g.

pow(3, 47159012670, 47159012671) is 1 # False
pow(3, 47159012670, 47159012671) == 1 # True

“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.”

Why? Because small integers are more frequently used by loops. Using reference to existing objects instead of creating a new object saves an overhead.

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    Just to make it clear: this is valid for the CPython interpreter. The language Python does not define this and other interpreters are free to have their own implementation. – Klaus D. Apr 19 at 5:07
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    Also 10e5 is a float, not an int. (Also, not all small ints come from the small int cache. For example, on current CPython, pow(3, 47159012670, 47159012671) == 1, but pow(3, 47159012670, 47159012671) is not 1.) – user2357112 Apr 19 at 5:37

If you take a look at Objects/longobject.c, which implements the int type for CPython, you will see that the numbers between -5 (NSMALLNEGINTS) and 256 (NSMALLPOSINTS - 1) are pre-allocated and cached. This is done to avoid the penalty of allocating multiple unnecessary objects for the most commonly used integers. This works because integers are immutable: you don't need multiple references to represent the same number.


Python doesn't know anything until you tell it. So in your code above, when you initialize a and b, you are storing those values(in the register or RAM), and calling the place to store it a and b, so that you can reference them later. If you didn't initialize the variable first, python would just give you an error.

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    I think you're missing the point of the question. a == b is obviously true. OP is asking why a is b is true. – Mad Physicist Apr 19 at 3:03

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