Why does untriggered assignment changed `nonlocal` behavior? [duplicate]

Question

This question already has an answer here:

• In Python, variables inside if conditions hide the global scope even if they are not executed? 4 answers

Today I'm reading python change log and meet the nonlocal keyword and did some experiment with it. I find a confusing situation where untriggered assignment will change the nonlocal keyword behavior, please see the example below.

def a():
    x = 'a'
    def b():
        def c():
            nonlocal x
            x = 'c'
        c()
    b()
    print(x)

a()

>>> python3 test.py
c


def a():
    x = 'a'
    def b():
        def c():
            nonlocal x
            x = 'c'
        c()
        if False:
            x = 'b'
    b()
    print(x)

a()

>>> python3 test2.py
a

You can saw that in test2.py, there is an untriggered assginment x = 'b' which changed the behavior of nonlocal.

Why this happened?

Answer 1

Python decides which variables are local to a function at compile time. x is assigned to within the function b, so it's local. That that branch is never actually reached at runtime is irrelevant and can't be decided in general.

So the x in c that is nonlocal is the next x in an outer scope, namely the one in b.

The alternative would be much more surprising -- consider what would happen if the if False: was instead if rand(10) == 6:. Then during the first call of b the nonlocal variable would refer to the outermost one, but randomly at some later call of b it would start referring to the other one!