Python 非本地和全局如何协同工作？

Question

This is my example

x=0
def outer():
    x = 1
    def i1():
        nonlocal x
        x = 2
        print("inner1:", x)

    i1()
    print("outer:", x)

    def i2():
        nonlocal x
        x = 3
        print("inner2:", x)

    i2()
    print("outer:", x)

    def i3():
        global x
        print("inner3:", x)

    i3()
    print("outer:", x)

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

Output in my Jupyter

inner1: 2
outer: 2
inner2: 3
outer: 3
inner3: 0
outer: 3
global: 0

Why does outer have value 0?

Answer 1

In i3() , when you declare global x , it indeed uses the outermost x , but you haven't changed its value.

In this part of the code:

i3()
print("outer:", x)

The print command is outside the i3() method, hence the global x is NOT being used. It is the local x which will be used. The global command in i3() means that only the x used within i3() will be global. Once outside i3() , the declared global scope for x will end.

Thus, print("outer:", x) prints 3, which is the outer() method's local variable's value. The outermost x remains 0 throughout.

Answer 2

I think your test case has a bug. If I change i3 to:

    def i3():
        global x
        x = "i3"
        print("inner3:", x)

then I get

global: i3

at the end, as I would expect.