为什么设置默认参数值会使此函数成为闭包？

Question

我正在编写一个应用程序,其中标签是可链接的,并且需要检索整个链接标签链.不允许自我引用.运行以下代码最终会得到一些非常奇怪的结果:

class Tag(object):
  def __init__(self, name):
    self.name = name
    self.links = []

  def __repr__(self):
    return "<Tag {0}>".format(self.name)

  def link(self, tag):
    self.links.append(tag)


def tag_chain(tag, known=[]):
  chain = []
  if tag not in known:
    known.append(tag)
  print "Known: {0}".format(known)

  for link in tag.links:
    if link in known:
      continue
    else:
      known.append(link)
    chain.append(link)
    chain.extend(tag_chain(link, known))
  return chain

a = Tag("a")
b = Tag("b")
c = Tag("c")
a.link(b)
b.link(c)
c.link(a)

o = tag_chain(a)
print "Result:", o
print "------------------"
o = tag_chain(a)
print "Result:", o

结果:

Known: [<Tag a>]
Known: [<Tag a>, <Tag b>]
Known: [<Tag a>, <Tag b>, <Tag c>]
Result: [<Tag b>, <Tag c>]
------------------
Known: [<Tag a>, <Tag b>, <Tag c>]
Result: []

所以,不知何故,我不小心创建了一个闭包.据我所见,已知应该已经超出范围并在函数调用完成后消失.

如果我将chain_tags()的定义更改为未设置默认值,则问题就会消失:

...
def tag_chain(tag, known):
...
o = tag_chain(a, [])
print "Result:", o
print "------------------"
o = tag_chain(a, [])
print "Result:", o

为什么是这样？

Answer 1

这是Python中常见的错误:

def tag_chain(tag, known=[]):
  # ...

known=[]并不意味着如果已知未被提及,则将其列为空列表; 事实上,它与"匿名"列表相关联.每次知道默认为该列表时,它都是相同的列表.

在这里做你想要的典型模式是:

def tag_chain(tag, known=None):
    if known is None:
        known = []
    # ...

known如果未提供,则正确初始化为空列表.