Python中nonlocal关键字

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转载最新推荐文章于 2024-02-08 13:00:28 发布 · 230 阅读

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原文链接：http://www.cnblogs.com/liubocoding/archive/2013/04/03/2997791.html

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本文详细介绍了Python中闭包与nonlocal关键字的使用方法，特别是nonlocal在Python3.0中的引入及其与Python2.x版本的区别。通过示例演示了在不同版本的Python中如何使用nonlocal关键字解决嵌套函数引用外部变量的问题，并提供了借助list或dict实现类似功能的替代方案。

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在这篇文章中介绍了“Python的闭包与nonlocal”的用法，因为nonlocal是在Python3.0中新增的关键字，python2.x不提供支持，文中提出在Python2.x解决嵌套函数引用外部变量的方法只有使用global 关键字定义全局变量，另一种可行的解决方案是使用列表或字典代替要操作的关键字。

示例如下：

1.python3.0 使用nonlocal关键字

>>> def outer():
        x = 1
        def inner():
            nonlocal x
            x = 2
            print("inner:", x)
        inner()
        print("outer:", x)

 >>> outer()
 inner: 2
 outer: 2

2.python2.x 借助list或dict实现

>>> def outer():
        x = [1]
        def inner():
            x[0] += 1 #修改x[0]保存的值
            print("inner:", x[0])
        inner()
        print("outer:", x[0])
 >>> outer()
 inner: 2
 outer: 2

2.javascript中作用域链定义与python作用域的异同

3.官方文档：

“The nonlocal statement causes the listed identifiers to refer to previously bound variables in the nearest enclosing scope. This is important because the default behavior for binding is to search the local namespace first. The statement allows encapsulated code to rebind variables outside of the local scope besides the global (module) scope.

Names listed in a nonlocal statement, unlike to those listed in a global statement, must refer to pre-existing bindings in an enclosing scope (the scope in which a new binding should be created cannot be determined unambiguously).

Names listed in a nonlocal statement must not collide with pre-existing bindings in the local scope.“

转载于:https://www.cnblogs.com/liubocoding/archive/2013/04/03/2997791.html