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