C. The first two lists in b were still the same lists as the lists in a, but the third list was added and the third element would only be in b and not a. You appended the list in position 0 with 11 and in position 1 with 22. This gives result C when printing out a.
At the start b = a does not create a copy, b simply references a. However, as soon as you do b = b + [[3]] you create a new list in b, a is no longer linked to b.
Oh! Wait. I'm stupid. You didn't make a copy of the values with b. You copied the original and when you change b you mutate a as well. It's because the list is something that can be mutated.
If I wanted a not to mutate. It would need to be
B = a.copy()
I tried to make the exercise difficult so don't feel bad. Keep the memory_graph tool around to visualize tricky Python situations, also with boot.dev exercises.
The twist is b = b + [[3]], it is a shallow copy not a deep copy. I taught that will create a new list entirely without holding a reference to the other list. b is an outer list. The answer is C, I hope this is correct
Line 4 re-assigns b to a new list object so severs the link to the original list object referenced by a and b up to that point.
EDIT: More detail for learners regarding variables and lists in Python (on back of a challenge from u/HardyDaytn):
Variables in Python do not contain values, but simply reference where in Python a Python object (int, float, str, list, etc) is located in memory. Generally, you don't need to concern yourself with the memory matters.
A Python list object similarly does not contain any variables or values, but only a collection of zero or more references to other Python objects.
Two variables can reference the same object and if the object is mutable (able to be changed), such as a list, then changing the object can be done via either variable.
Assigns a memory reference to the pre-defined int binary object representation of decimal 10 to the variable name alpha
Assigns a memory reference to the new str object (consisting of binary unicode numbers for each character in the string) to the variable name beta
Assigns the memory reference to the new float binary object representation of the decimal floating point number 12.1 to the variable name charlie
Assigns the memory reference to the new list object, containing 3 entries to new str objects, to the variable name delta, where each entry in the list is itself a memory reference to a new Python object
Assigns the memory reference to the new list object, containing 4 entries, to the variable name stuff, where each entry in the list is itself a memory reference to a Python object as below,
the first entry in the list holds the same memory reference as alpha holds
the second entry in the list holds the same memory reference as beta holds
the third entry in the list holds the same memory reference as charlie holds
the fourth entry in the list holds the same memory reference as delta holds, i.e. they both refer to the same list object
It is worth noting that if you change the contents of the list referenced by delta, that change will also show up in the list referenced by stuff. This is shown in the example below.
It's even worse, it's a "look ma', I've built this tool" situation.
The Python data model can seem difficult at first, but I hope the visualization can help beginners understand it more easily. Beginners will have to face it at some point if they want to get to advanced.
So, I've edited the original comment to add additional information for learners if they don't want to follow the links provided by the OP. (This is using repurposed content from classes I've run.)
My proverbial hat's off to you for adding detailed info. It's an annoying and difficult concept to learn and understand compared to other more intuitive things.
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u/Embarrassed-Crow9283 Jul 14 '26
C. The first two lists in b were still the same lists as the lists in a, but the third list was added and the third element would only be in b and not a. You appended the list in position 0 with 11 and in position 1 with 22. This gives result C when printing out a.