Question

What does nonlocal do in Python 3.x?

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_{To close debugging questions where OP needs nonlocal and doesn’t realize it, please use Is it possible to modify variable in python that is in outer, but not global, scope? instead.}

_{Although Python 2 is officially unsupported as of January 1, 2020, if for some reason you are forced to maintain a Python 2.x codebase and need an equivalent to nonlocal, see nonlocal keyword in Python 2.x.}

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