Python，如何在父范围内更改变量的值? [英] Python, how can I change value of a variable in the parent scope?

问题描述

例如: assginment语句将声明一个新的局部变量.

for example: assginment statement will declare a new local variable.

foo = 'global'
def func1():
    foo = 'func1'
    def func2():
        foo = 'local variable in func2'

use global声明将在global中使用foo:

use global declaration will use the foo in global:

def func2():
    global foo
    foo = 'global changed in func2'  #changed the foo value in global scope

如何在func1范围内更改变量foo?
感谢您的帮助.

how can I change the variable foo in func1 scope?
Thanks for any help.

谢谢布兰登·克雷格·罗兹，我终于明白了你的意思.

Thank you Brandon Craig Rhodes, I finally understand your meaning.

如果嵌套了三个以上的作用域，则可以将变量存储在列表中.

if there are more than 3 scopes nested, I can store the variable in a list.

foo = ['global', 'function1', 'function2']
def func1():
    foo[1] = 'func1'
    def func2():
        foo[2] = 'func2'
        foo[1] = 'func1 modified in func2'

我实际上只是使用一个全局变量.

I just use a global variable actually.

因此，如果嵌套了两个函数，我们可以使用

so, if there are two functions nested, we can use

nonlocal foo 

和

global foo 

如果嵌套了三个以上的函数，则
并且每个函数在其他函数范围内使用变量，
为什么不声明全局列表变量?
感谢您的所有帮助！

if there are more than three functions nested,
and each function use variables in other functions scope,
why don't we declare a global list variable?
Thank you for all your help!!!

推荐答案

我相信在Python 3中，您可以使用nonlocal关键字来获得在非全局范围内修改变量的权限.在Python 2中，您不能在封闭范围内重新分配foo.而是将foo设置为等于诸如列表[]的可变对象，然后将要存储的值粘贴在列表中:

In Python 3, I believe, you can use the nonlocal keyword to get permission to modify a variable in an enclosing non-global scope. In Python 2, you cannot reassign foo in an enclosing scope; instead, set foo equal to a mutable object like a list [] and then stick the value you want stored in the list:

def func1():
    foo = [None]
    def func2():
        foo[0] = 'Test'

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