I wanted to make sense of the following code variations:
a = [1, 2, 3]
b = a
b.append(4)
b = ['a', 'b']
print(a, b)
What I understood was that variable a refers to an object that contains the list [1,2,3] in some place in memory, and b now is referring to the same object that a is referring to, and via that link we're technically appending in a not b.
Output: [1, 2, 3, 4] ['a', 'b']
I updated the code a bit:
a = [1, 2, 3]
b = ['a', 'b']
b = a
b.append(4)
print(a, b)
My understanding: b is now referring to two objects, the first list ['a','b'] and the second list (that a is initially referring to) [1,2,3] via the third line b = a.
Output: [1, 2, 3, 4] [1, 2, 3, 4]
Last Code variation:
a = [1, 2, 3]
b = ['a', 'b']
b = a
b.append(4)
a.append(10)
print(a, b)
based on my understanding so far, I though that the link on line 3 b = a was giving only b the ability to reference multiple objects (it's own and a's) and a should've only be referencing one object [1,2,3], so the expected output should be: [1,2,3,4,10] [1,2,3,4]
Actual Output: [1, 2, 3, 4, 10] [1, 2, 3, 4, 10]
So is this assignment on line 3 b = a is like a bi-directional link? where also a reference is created by a to b's object?
Don't think of the objects like pointers, I think that is the source of your confusion. It is not that "b points to a" or "a points to b", it has to do with binding to an object. I think looking at id will be useful
>>> a = [1, 2, 3]
>>> id(a)
1833964774216
>>> b = a
>>> id(b)
1833964774216
In this case both a and b are bound to that list. So any mutation to list 1833964774216 will be reflected in both objects. But I can re-assign (or re-bind) to a completely different object
>>> b = [4, 5, 6]
>>> id(b)
1833965089992
>>> b
[4, 5, 6]
This has no effect whatsoever on a because it is still bound to the original list
>>> a
[1, 2, 3]
>>> id(a)
1833964774216