为什么当最后一个强引用消失时，没有从 WeakValueDictionary 中删除值

Question

我有以下 Python 程序：

import weakref

class NumberWord:
  def __init__(self, word):
    self.word = word
  def __repr__(self):
    return self.word

dict = weakref.WeakValueDictionary()

print(f"[A] {len(dict)}")
print(f"dict.get(0) = {dict.get(0)}")
print(f"dict.get(1) = {dict.get(1)}")

list = []
list.append(NumberWord("zero"))
dict[0] = list[0]

print(f"[B] {len(dict)}")
print(f"dict.get(0) = {dict.get(0)}")
print(f"dict.get(1) = {dict.get(1)}")

list.append(NumberWord("one"))
dict[1] = list[1]
print(list)

print(f"[C] {len(dict)}")
print(f"dict.get(0) = {dict.get(0)}")
print(f"dict.get(1) = {dict.get(1)}")

list.pop()
print(list)

print(f"[D] {len(dict)}")
print(f"dict.get(0) = {dict.get(0)}")
print(f"dict.get(1) = {dict.get(1)}")

list.pop()
print(list)

print(f"[E] {len(dict)}")
print(f"dict.get(0) = {dict.get(0)}")
print(f"dict.get(1) = {dict.get(1)}")

我预计会出现以下行为：

在第 [A] 步，字典为空
在步骤 [B] 中，字典包含 dict[0] = NumberWord("zero")
在步骤 [C] 中，字典包含 dict[0] = NumberWord("zero") 和 dict[1] = NumberWord("one")
在步骤 [D] 中，字典包含 dict[1] = NumberWord("one")（"zero" 被删除，因为列表中唯一的强引用消失了）
在步骤 [E] 中，dict 再次为空（"one" 被删除，因为列表中唯一的强引用消失了）

一切正常除了步骤 [E]："one" 不会消失。为什么不呢？

这里是实际输出：

>>> import weakref
>>> 
>>> class NumberWord:
...   def __init__(self, word):
...     self.word = word
...   def __repr__(self):
...     return self.word
... 
>>> dict = weakref.WeakValueDictionary()
>>> 
>>> print(f"[A] {len(dict)}")
[A] 0
>>> print(f"dict.get(0) = {dict.get(0)}")
dict.get(0) = None
>>> print(f"dict.get(1) = {dict.get(1)}")
dict.get(1) = None
>>> 
>>> list = []
>>> list.append(NumberWord("zero"))
>>> dict[0] = list[0]
>>> 
>>> print(f"[B] {len(dict)}")
[B] 1
>>> print(f"dict.get(0) = {dict.get(0)}")
dict.get(0) = zero
>>> print(f"dict.get(1) = {dict.get(1)}")
dict.get(1) = None
>>> 
>>> list.append(NumberWord("one"))
>>> dict[1] = list[1]
>>> print(list)
[zero, one]
>>> 
>>> print(f"[C] {len(dict)}")
[C] 2
>>> print(f"dict.get(0) = {dict.get(0)}")
dict.get(0) = zero
>>> print(f"dict.get(1) = {dict.get(1)}")
dict.get(1) = one
>>> 
>>> list.pop()
one
>>> print(list)
[zero]
>>> 
>>> print(f"[D] {len(dict)}")
[D] 2
>>> print(f"dict.get(0) = {dict.get(0)}")
dict.get(0) = zero
>>> print(f"dict.get(1) = {dict.get(1)}")
dict.get(1) = one
>>> 
>>> list.pop()
zero
>>> print(list)
[]
>>> 
>>> print(f"[E] {len(dict)}")
[E] 1
>>> print(f"dict.get(0) = {dict.get(0)}")
dict.get(0) = zero
>>> print(f"dict.get(1) = {dict.get(1)}")
dict.get(1) = None
>>> 
>>>

Answer 1

我自己找到了答案。

原因是特殊变量_仍然包含上次计算的结果。

上次评价为list.pop()，结果为NumberWord("zero")。

只要此结果仍存储在 _ 中，我们就会继续拥有强引用，而弱引用不会消失。

我们可以通过进行另一次评估来证实这一理论。那时 _ 将包含一个不同的值，弱引用将消失：

如果我们在上面例子的末尾执行下面的附加语句：

_
5 + 5
_
print(f"[F] {len(dict)}")
print(f"dict.get(0) = {dict.get(0)}")
print(f"dict.get(1) = {dict.get(1)}")

然后我们得到如下输出：

>>> _
zero
>>> 5 + 5
10
>>> _
10
>>> print(f"[F] {len(dict)}")
[F] 0
>>> print(f"dict.get(0) = {dict.get(0)}")
dict.get(0) = None
>>> print(f"dict.get(1) = {dict.get(1)}")
dict.get(1) = None

为什么当最后一个强引用消失时，没有从 WeakValueDictionary 中删除值

Why is value not removed from WeakValueDictionary when last stong reference goes away

python

dictionary

weak-references