Python scope of int vs scope of list

Question

I am having troubles understanding the scope concept in Python. When I use the following code:|

x = 100

def printer(x):
    x = 25
    print(x)

printer(x) -> prints 25
print(x) -> prints 100

In the example above I understand the variable 'x' only has 25 as value in the function. When I try to do the same with a list, my output is different. See code below:

board = ['','']

def printer():
    board[0] = 'X'
    print(board)

printer() --> prints ['X', '']
print(board) --> also prints ['X', ''] || Here I was expecting to print ['','']

This has **nothing** to do with scope. In one case, you are *re-assigning the variable `x`, in the other, your are *mutating the object referenced by the variable x*. Note, change `board = ['X', '']` in your second example and you'll see it works the same. — juanpa.arrivillaga, Jan 08 '20 at 12:35
Read the following: https://nedbatchelder.com/text/names.html — juanpa.arrivillaga, Jan 08 '20 at 12:36
Lists are working with referance so if you change it in a function then it will change on local — ikibir, Jan 08 '20 at 12:36
@ikibir `list` and `int` objects both *work exactly the same*, they *both* use reference semantics, `int` objects *merely lack mutator methods*. — juanpa.arrivillaga, Jan 08 '20 at 12:37
It's somewhat about scope, as the change to `board = ['X', '']` changes `board` from a free variable to a local variable. — chepner, Jan 08 '20 at 12:39

score 5 · Answer 1 · answered Jan 08 '20 at 12:36

Only names have scopes. The x you assign before printer is a global name. The parameter x in printer is local to printer.

board is a global name. Since you never assign to board inside printer (board[0] = 'X' is not an assignment to board; it's a method call board.__setattr__(0, 'X')), it is a free variable whose value is taken from a name in an enclosing scope. The assignment mutates an existing value referenced by the global name board.

Python scope of int vs scope of list

1 Answers1