相思资源网 Design By www.200059.com
搞不清楚在闭包(closures)中Python是怎样绑定变量的
看这个例子:
> def create_multipliers(): ... return [lambda x : i * x for i in range(5)] > for multiplier in create_multipliers(): ... print multiplier(2) ...
期望得到下面的输出:
0
2
4
6
8
但是实际上得到的是:
8
8
8
8
8
实例扩展:
# coding=utf-8 __author__ = 'xiaofu' # 解释参考 http://docs.python-guide.org/en/latest/writing/gotchas/#late-binding-closures def closure_test1(): """ 每个closure的输出都是同一个i值 :return: """ closures = [] for i in range(4): def closure(): print("id of i: {}, value: {} ".format(id(i), i)) closures.append(closure) # Python's closures are late binding. # This means that the values of variables used in closures are looked up at the time the inner function is called. for c in closures: c() def closure_test2(): def make_closure(i): def closure(): print("id of i: {}, value: {} ".format(id(i), i)) return closure closures = [] for i in range(4): closures.append(make_closure(i)) for c in closures: c() if __name__ == '__main__': closure_test1() closure_test2()
输出:
id of i: 10437280, value: 3 id of i: 10437280, value: 3 id of i: 10437280, value: 3 id of i: 10437280, value: 3 id of i: 10437184, value: 0 id of i: 10437216, value: 1 id of i: 10437248, value: 2 id of i: 10437280, value: 3
标签:
Python,闭包,绑定变量
相思资源网 Design By www.200059.com
广告合作:本站广告合作请联系QQ:858582 申请时备注:广告合作(否则不回)
免责声明:本站文章均来自网站采集或用户投稿,网站不提供任何软件下载或自行开发的软件! 如有用户或公司发现本站内容信息存在侵权行为,请邮件告知! 858582#qq.com
免责声明:本站文章均来自网站采集或用户投稿,网站不提供任何软件下载或自行开发的软件! 如有用户或公司发现本站内容信息存在侵权行为,请邮件告知! 858582#qq.com
相思资源网 Design By www.200059.com
暂无Python新手如何进行闭包时绑定变量操作的评论...