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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转载请注明原文链接：Python新手如何进行闭包时绑定变量操作
