1

If I use the below:

a = 1000
print(id(a))

myList = [a,2000,3000,4000]
print(id(myList[0]))

# prints the same IDs

I get the same id. This makes sense to me. I can understand how the memory manager could assign the same object to these variables, because I am directly referencing a in the list.

However, if I do this instead:

a = 1000
print(id(a))

myList = [1000,2000,3000,4000]
print(id(myList[0]))

# prints the same IDs

I STILL get the same id being output for both prints. How does Python know to use the same object for these assignments? Searching for pre-existence would surely be hugely inefficient so I am presuming something more clever is going on here.

My first thought was something to do with the integer itself being used to calculate the objects address, but the behaviour also holds true for strings:

a = "car"
print(id(a))

myList = ["car",2000,3000,4000]
print(id(myList[0]))

# prints the same IDs

The behaviour does NOT however, hold true for list elements:

a = [1,2,3]
print(id(a))

myList = [[1,2,3],2000,3000,4000]
print(id(myList[0]))

# prints different IDs

Can someone explain the behaviour I am seeing?

EDIT - I have encountered that for small values between -5 and 256, the same object may be used. The thing is that I am seeing the same object still being used even for huge values, or even strings:

a = 1000000000000
myList = [1000000000000,1000,2000]
print(a is myList[0])
# outputs True!

My question is How can Python work out that it is the same object in these cases without searching for pre-existence? Let's say CPython specifically

EDIT - I am using Python V3.8.10

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  • Does this answer your question? "is" operator behaves unexpectedly with integers Commented Jan 30, 2023 at 13:56
  • @Homer512 not particularly The same object is used here even if the number is huge, not just between -5 and 256 Commented Jan 30, 2023 at 13:58
  • for immutable object it store them in heap and map the variable to those heap object, if any variable want to set same value then from heap it give back the same address to the variable if that value in heap , and thus variable gives same id Commented Jan 30, 2023 at 14:03
  • Can you also update your question with the python version you are using? Commented Jan 30, 2023 at 14:06
  • 1
    Maybe stackoverflow.com/q/34147515/12671057 Commented Jan 30, 2023 at 14:11

1 Answer 1

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In Python, small and unchanging values like numbers and short strings are stored only once (unless operators are used for making them, this way a new object is created for that) in the computer's memory to save space and speed up the program. This process is called "interning". This means that when you write the same value multiple times, it will have the same memory address (id), and you will get the same id for each instance of that value. However, lists and other more complex data types are not interned, so every time you use a list, a new memory space is allocated for it, giving it a different id.

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4 Comments

@KellyBundy When you change the value of a variable using an operation, such as y = -x, a new object is created to store the new value, even if it's the same as the original value. This new object will have a different memory address (id) from the original variable, so x and y have different ids even though their values are the same.
@KellyBundy This is because the operation --x creates a new object in memory, so even though it has the same value as x, it is stored in a different memory location and has a different id. In Python, when an operator is applied to an object, a new instance of that object may be created, resulting in a different id.
As I said, using operators ALWAYS makes a new object, although they have the same value. I changed my answer with more clarification. Thanks.
Interning is not universal or defined by Python. It's something that CPython (and other implementations) can do.

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