Why is the id of a Python class not unique when called quickly?

Question

I'm doing some things in Python (3.3.3), and I came across something that is confusing me since to my understanding classes get a new id each time they are called.

Lets say you have this in some .py file:

class someClass: pass

print(someClass())
print(someClass())

The above returns the same id which is confusing me since I'm calling on it so it shouldn't be the same, right? Is this how Python works when the same class is called twice in a row or not? It gives a different id when I wait a few seconds but if I do it at the same like the example above it doesn't seem to work that way, which is confusing me.

>>> print(someClass());print(someClass())
<__main__.someClass object at 0x0000000002D96F98>
<__main__.someClass object at 0x0000000002D96F98>

It returns the same thing, but why? I also notice it with ranges for example

for i in range(10):
    print(someClass())

Is there any particular reason for Python doing this when the class is called quickly? I didn't even know Python did this, or is it possibly a bug? If it is not a bug can someone explain to me how to fix it or a method so it generates a different id each time the method/class is called? I'm pretty puzzled on how that is doing it because if I wait, it does change but not if I try to call the same class two or more times.

Answer 1

The id of an object is only guaranteed to be unique during that object's lifetime, not over the entire lifetime of a program. The two someClass objects you create only exist for the duration of the call to print - after that, they are available for garbage collection (and, in CPython, deallocated immediately). Since their lifetimes don't overlap, it is valid for them to share an id.

It is also unsuprising in this case, because of a combination of two CPython implementation details: first, it does garbage collection by reference counting (with some extra magic to avoid problems with circular references), and second, the id of an object is related to the value of the underlying pointer for the variable (ie, its memory location). So, the first object, which was the most recent object allocated, is immediately freed - it isn't too surprising that the next object allocated will end up in the same spot (although this potentially also depends on details of how the interpreter was compiled).

If you are relying on several objects having distinct ids, you might keep them around - say, in a list - so that their lifetimes overlap. Otherwise, you might implement a class-specific id that has different guarantees - eg:

class SomeClass:
    next_id = 0

    def __init__(self):
         self.id = SomeClass.nextid
         SomeClass.nextid += 1