Python闭包时绑定变量实例 Python新手怎样进行闭包时绑定变量操作
爱喝马黛茶的安东尼 人气:0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
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