Python enclosing scope: at call point or at definition?

Asked Sep 12 '19 at 03:57

Active Sep 12 '19 at 04:07

Viewed 56 times

This is an example of an enclosing scope (is it also called lexical scope?):

def outer(a):
    def inner():
        return a
    return inner

if __name__ == '__main__':
    inner = outer(10)
    print(inner())

This code also works:

def printer():
    print(x)

if __name__ == '__main__':
    x = 10
    printer()

My question is: Is the "enclosing scope" rule used in the second example? If so, is there any distinction between the 2 types of enclosing scope: at call point/runtime and at definition?

Edit: as @user2357112 pointed out, x is global. This one doesn't work:

def printer():
    print(x)

def outer(func_print):
    x = 10
    func_print()


if __name__ == '__main__':
    outer(printer)

What a rookie mistake...

edited Sep 12 '19 at 04:07

asked Sep 12 '19 at 03:57

Minh Nghĩa

2

That `x` is a global. – user2357112 Sep 12 '19 at 04:00
@user2357112 Thanks, I'm so embarrassed. – Minh Nghĩa Sep 12 '19 at 04:08
as soon as you *assign* to a name inside a function it becomes local (and shadows the global name inside the function). if you wanted to change the global `x` in your last example you'd have to use `global x; x = 10`. – hiro protagonist Sep 12 '19 at 04:20
also this might help: https://stackoverflow.com/a/292502/4954037 – hiro protagonist Sep 12 '19 at 04:24

Python enclosing scope: at call point or at definition?

0 Answers0