情况一： a 直接引用外部的，正常运行

def toplevel():

    a = 5

    def nested():

        print(a + 2) # theres no local variable a so it prints the nonlocal one

    nested()

    return a

情况二：创建local 变量a，直接打印，正常运行

def toplevel():

    a = 5

    def nested():

        a = 7 # create a local variable called a which is different than the nonlocal one

        print(a) # prints 7

    nested()

    print(a) # prints 5

    return a

情况三：由于存在 a = 7，此时a代表嵌套函数中的local a , 但在使用a + 2 时，a还未有定义出来，所以报错

def toplevel():

    a = 5

    def nested():

        print(a + 2) # tries to print local variable a but its created after this line so exception is raised

        a = 7

    nested()

    return a

toplevel()

针对情况三的解决方法, 在嵌套函数中增加nonlocal a ,代表a专指外部变量即可

def toplevel():

    a = 5

    def nested():

        nonlocal a

        print(a + 2)

        a = 7

    nested()

    return a