Python覆盖嵌套函数中的变量 [英] Python overwriting variables in nested functions

问题描述

假设我有以下 python 代码:

Suppose I have the following python code:

def outer():
    string = ""
    def inner():
        string = "String was changed by a nested function!"
    inner()
    return string

我想调用outer()来返回字符串被嵌套函数改变了！"，但我得到了".我的结论是，Python 认为 string = "string was changed by an nested function!" 行是对 inner() 局部新变量的声明.我的问题是:我如何告诉 Python 它应该使用 outer() 字符串?我不能使用 global 关键字，因为该字符串不是全局的，它只是存在于外部作用域中.想法?

I want a call to outer() to return "String was changed by a nested function!", but I get "". I conclude that Python thinks that the line string = "string was changed by a nested function!" is a declaration of a new variable local to inner(). My question is: how do I tell Python that it should use the outer() string? I can't use the global keyword, because the string isn't global, it just lives in an outer scope. Ideas?

推荐答案

在 Python 3.x 中，您可以使用 nonlocal 关键字:

In Python 3.x, you can use the nonlocal keyword:

def outer():
    string = ""
    def inner():
        nonlocal string
        string = "String was changed by a nested function!"
    inner()
    return string

在 Python 2.x 中，您可以使用包含单个元素的列表并覆盖该单个元素:

In Python 2.x, you could use a list with a single element and overwrite that single element:

def outer():
    string = [""]
    def inner():
        string[0] = "String was changed by a nested function!"
    inner()
    return string[0]

这篇关于Python覆盖嵌套函数中的变量的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！