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)}")
我希望以下行为:
在步骤 [A] 中,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")
("零"被删除,因为列表中唯一的强引用消失了(在步骤 [E] 中,字典再次为空("一"被删除,因为列表中唯一的强引用消失了(
除步骤 [E]外,一切都按预期工作:"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
>>>
>>>
我自己刚刚发现了答案。
原因是特殊变量_
它仍然包含上次评估的结果。
最后一次评估是list.pop()
,其结果是NumberWord("zero")
。
只要这个结果仍然存储在_
我们就会继续拥有强引用,并且弱引用不会消失。
我们可以通过进行另一次评估来证实这一理论。此时_
将包含一个不同的值,弱引用将消失:
如果我们在上面示例的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