它取源代码中“最接近”的引用点。 这被称为“词法范围”，是>40年来的标准。

Python的类成员实际上在一个名为__dict__的字典中，并且永远不会通过词法作用域到达。

如果你不指定nonlocal，而是指定x = 7，它将创建一个新的局部变量"x"。 如果你指定了nonlocal，它会找到“最接近”的“x”并赋值给它。 如果你指定了nonlocal并且没有“x”，它会给你一个错误消息。

关键字global对我来说总是很奇怪，因为它会很高兴地忽略所有其他的“x”，除了最外面的那个。

引用自Python 3

非局部语句导致列出的标识符引用之前在最近的封闭范围内绑定的变量(不包括全局变量)。

如参考文献中所述，在多个嵌套函数的情况下，只有最近的封闭函数中的变量被修改:

def outer():
    def inner():
        def innermost():
            nonlocal x
            x = 3

        x = 2
        innermost()
        if x == 3: print('Inner x has been modified')

    x = 1
    inner()
    if x == 3: print('Outer x has been modified')

x = 0
outer()
if x == 3: print('Global x has been modified')

# Inner x has been modified

“最近的”变量可能在几层之外:

def outer():
    def inner():
        def innermost():
            nonlocal x
            x = 3

        innermost()

    x = 1
    inner()
    if x == 3: print('Outer x has been modified')

x = 0
outer()
if x == 3: print('Global x has been modified')

# Outer x has been modified

但它不能是一个全局变量:

def outer():
    def inner():
        def innermost():
            nonlocal x
            x = 3

        innermost()

    inner()

x = 0
outer()
if x == 3: print('Global x has been modified')

# SyntaxError: no binding for nonlocal 'x' found

比较一下，不使用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

为此，使用非局部，其中inner()的x现在也是outer()的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

如果我们要使用global，它会将x绑定到正确的“global”值:

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

简而言之，它允许您在外部(但非全局)作用域中为变量赋值。参阅PEP 3104了解所有血腥的细节。

文件如下:

非局部语句导致所列出的标识符被引用 先前在最近的封闭范围内绑定的变量除外 全局变量。

例如，inner()中的非局部变量foo可以访问middle()中的非局部变量foo = 10，但不能访问outer()中的非局部变量foo = 5或outer()外的全局变量foo = 0，如下所示:

foo = 0 # <- ✖
def outer():
    foo = 5 # <- ✖
    def middle():
        foo = 10 # <- 〇
        def inner():
            nonlocal foo # Here
            foo += 1
            print(foo) # 11
        inner()
    middle()
outer()

help('nonlocal') The nonlocal statement nonlocal_stmt ::= "nonlocal" identifier ("," identifier)* The nonlocal statement causes the listed identifiers to refer to previously bound variables in the nearest enclosing scope. This is important because the default behavior for binding is to search the local namespace first. The statement allows encapsulated code to rebind variables outside of the local scope besides the global (module) scope. Names listed in a nonlocal statement, unlike to those listed in a global statement, must refer to pre-existing bindings in an enclosing scope (the scope in which a new binding should be created cannot be determined unambiguously). Names listed in a nonlocal statement must not collide with pre- existing bindings in the local scope. See also: PEP 3104 - Access to Names in Outer Scopes The specification for the nonlocal statement. Related help topics: global, NAMESPACES

来源:Python语言参考