Messages in this thread | | | Date | Fri, 31 Jan 2020 16:43:09 +0000 | From | Will Deacon <> | Subject | Confused about hlist_unhashed_lockless() |
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Hi folks,
I just ran into c54a2744497d ("list: Add hlist_unhashed_lockless()") but I'm a bit confused about what it's trying to achieve. It also seems to have been merged without any callers (even in -next) -- was that intentional?
My main source of confusion is the lack of memory barriers. For example, if you look at the following pair of functions:
static inline int hlist_unhashed_lockless(const struct hlist_node *h) { return !READ_ONCE(h->pprev); }
static inline void hlist_add_before(struct hlist_node *n, struct hlist_node *next) { WRITE_ONCE(n->pprev, next->pprev); WRITE_ONCE(n->next, next); WRITE_ONCE(next->pprev, &n->next); WRITE_ONCE(*(n->pprev), n); }
Then running these two concurrently on the same node means that hlist_unhashed_lockless() doesn't really tell you anything about whether or not the node is reachable in the list (i.e. there is another node with a next pointer pointing to it). In other words, I think all of these outcomes are permitted:
hlist_unhashed_lockless(n) n reachable in list 0 0 (No reordering) 0 1 (No reordering) 1 0 (No reordering) 1 1 (Reorder first and last WRITE_ONCEs)
So I must be missing some details about the use-case here. Please could you enlighten me? The RCU implementation permits only the first three outcomes afaict, why not use that and leave non-RCU hlist as it was?
Cheers,
Will
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