Python overwriting variables in nested functions

python, scope

Solution

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

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]

Problem

Suppose I have the following python code: ``` def outer(): string = "" def inner(): string = "String was changed by a nested function!" inner() return string ``` 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?

Original source

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