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Date:   Fri, 31 Jan 2020 16:43:09 +0000
From:   Will Deacon <will@kernel.org>
To:     edumazet@google.com, tglx@linutronix.de, paulmck@kernel.org
Cc:     x86@kernel.org, linux-kernel@vger.kernel.org, peterz@infradead.org
Subject: Confused about hlist_unhashed_lockless()
Message-ID: <20200131164308.GA5175@willie-the-truck>
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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