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Python新手如何进行闭包时绑定变量操作
站长资源 2024/11/25 佚名
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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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