Assign a list variable to a new variable results in two variables with the SAME list. Why.

Question

Why this code results in appending a value to list a? So print(a) function outputs [2,4,6,8,5], while 5 was appended to list b?

a =[2,4,6,8]
b = a
b.append(5)
print(a)

Answer 1

I explain it in detail here (meant for reading - run it for seeing the examples):

https://code.sololearn.com/c89ejW97QsTN/?ref=app

Answer 2

PholaX, yeah, that then becomes a practical problem:

How can you write that function that it's the least likely to do damage?

Handing over the object itself (the reference to it) gives the function quite a bit of power.

So often it can be better to not hand over the... let's say list, but just have the function return the values *for* the list.

Or, if a list has to be worked on, let the function work on a copy of the list.

Do you have an example in mind? Might be interesting to consider the options.

Answer 3

Thank you, that was really helpful.
So in Python, names of variables are just references to data in memory and by simple assignment we only get a new reference with another name. Therefore to create a new... hmm... "container" in memory we must explicitly say we want a new "object" using the constructor (or by assigning a new unique value).
Right?

Answer 4

PholaX, exactly.

At least with mutable objects like lists and such we always need to know if we are refering to a similar object or that very object.

Answer 5

That's pretty hard, for example, when you have to pass a list as an argument to a function... But I also can see how it can be useful.
Thank you again.