Python nonlocal statement

Compare this, without using nonlocal:

x = 0
def outer():
    x = 1
    def inner():
        x = 2
        print("inner:", x)

    inner()
    print("outer:", x)

outer()
print("global:", x)

# inner: 2
# outer: 1
# global: 0

To this, using nonlocal, where inner()'s x is now also outer()'s x:

x = 0
def outer():
    x = 1
    def inner():
        nonlocal x
        x = 2
        print("inner:", x)

    inner()
    print("outer:", x)

outer()
print("global:", x)

# inner: 2
# outer: 2
# global: 0

    If we were to use global, it would bind x to the properly "global" value:

    x = 0
    def outer():
        x = 1
        def inner():
            global x
            x = 2
            print("inner:", x)

        inner()
        print("outer:", x)

    outer()
    print("global:", x)

    # inner: 2
    # outer: 1
    # global: 2