closures - Python nonlocal statement

Question

What does the Python nonlocal statement do (in Python 3.0 and later)?

There's no documentation on the official Python website and help("nonlocal") does not work, either.

Answer 1

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