问题
The following makes sense to me:
>>> [] is []
False
Given that lists are mutable, I would expect [] to be a new empty list object every time it appears in an expression. Using this explanation however, the following surprises me:
id([]) == id([])
True
Why? What is the explanation?
回答1:
In the first example, [] is not [] precisely because the lists are mutable. If they weren't, they could safely map to the same one without issue.
In the second example, id([]) creates a list, gets the id, and deallocates the list. The second time around it creates a list again, but "puts it in the same place" because nothing much else has happened. id is only valid during an object's lifetime, and in this case its lifetime is virtually nil
From the docs on id:
This is an integer (or long integer) which is guaranteed to be unique and constant for this object during its lifetime. Two objects with non-overlapping lifetimes may have the same id() value.
Commented disassembly:
0 LOAD_GLOBAL 0 (id) # load the id function
3 BUILD_LIST 0 # create the first list
6 CALL_FUNCTION 1 # get the id
9 LOAD_GLOBAL 0 (id) # load the id function
12 BUILD_LIST 0 # create the second list
15 CALL_FUNCTION 1 # get the id
18 COMPARE_OP 2 (==) # compare the two ids
21 RETURN_VALUE # return the comparison
Note there is no STORE_FAST to retain the list. Therefore it was discarded immediately after getting passed to the id function.
来源：https://stackoverflow.com/questions/21948712/when-does-python-create-new-list-objects-for-empty-lists