当最后一个强引用消失时，为什么不从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)}")

我期望以下行为：

• 在步骤[A]中，dict为空

• 在步骤[B]中，dict包含dict[0] = NumberWord("zero")

• 在步骤[C]中，dict包含dict[0] = NumberWord("zero")和dict[1] = NumberWord("one")

• 在步骤[D]中，dict包含dict[1] = NumberWord("one")（“零”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（“一”被删除，因为列表中唯一的强引用消失了）

除了步骤[E]：“一”不会消失。为什么不呢

以下是实际输出：

>>> 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

我自己才发现答案

原因是特殊变量_仍然包含上次计算的结果

最后一次评估是list.pop()，结果是NumberWord("zero")

只要这个结果仍然存储在_中，我们将继续使用强引用，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。此时_将包含不同的值，弱引用将消失：

如果我们在上述示例末尾执行以下附加语句：

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

然后我们得到以下输出：

>>> _
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