python--nolocal

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