非本地关键字似乎无效

Question

我希望foo内的变量发生变化，因为我在方法栏中使用了nonlocal关键字。我的理解是变量a，b，c不再是bar中的局部变量，它们只是foo中变量a，b，c的赋值，因此值应该在foo中更改。 我在python3中运行以下代码时没有看到这种情况。想知道我在想什么

下面是输出

From foo
a = 10
b = 20
c = 30

From bar
a = 11
b = 12
c = 19

From chu
a = 200
b = 300
c = 500

这是代码

def foo():
    a = 10
    b = 20
    c = 30
    print("From {func}\na = {a}\nb = {b}\nc = {c}\n".format(func="foo", a=a, b=b, c=c))

    def bar():
        nonlocal a
        nonlocal b
        nonlocal c
        a = 11
        b = 12
        c = 19
        print("From {func}\na = {a}\nb = {b}\nc = {c}\n".format(func="bar", a=a, b=b, c=c))

        def chu():
            a = 200
            b = 300
            c = 500
            print("From {func}\na = {a}\nb = {b}\nc = {c}\n".format(func="chu",a=a,b=b,c=c))
        chu()
    bar()
foo()