Python 闭包变量绑定问题

延时绑定

原文

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.

即闭包中变量的绑定发生在闭包被调用时, 而非闭包定义时.

实例

# 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

需要注意的是, 对于第二个例子closure_test2(), 在调用完make_closure(i)(i < 3)之后, 循环里的i值改变了, 然而make_closure()函数中的i是不会受到影响的, 这一点从输出的id值也可以看出.