Question

我试图提出一个简洁的示例，说明对象的垃圾回收，没有变量名持有对该对象的引用，但是此代码似乎不起作用。我想了解为什么更好地了解Python的内部功能。似乎暴露了我误会的东西。

some_name = [['some_nested_lst_object_with_an_str_object']]
id(some_name)
'''
you don't normally need to do this.
This is done for the reference example.
Accessing garbage collector:
'''
import gc
print(gc.collect())
'''
If I assign something new to ''*some_name*'',
the reference to the previous object will be lost:
'''
some_name
print(gc.collect())
some_name = [[['something_new']]]
some_name
print(gc.collect())

Answer 1

Python通常使用引用计数来释放对象。仅在循环引用的情况下，才需要垃圾回收：

some_name = [123]
print(gc.collect())
some_name = [] # previous some_name-object is freed directly
some_name.append(some_name) # cyclic reference
print(gc.collect()) # 0
some_name = None
print(gc.collect()) # 1

为什么这不会产生预期的垃圾？

1 个答案: