Re: [BUG]locking/rwsem: only clean RWSEM_FLAG_HANDOFF when already set

From: Waiman Long
Date: Thu Nov 11 2021 - 16:09:43 EST


On 11/11/21 15:50, Peter Zijlstra wrote:
So I suspect that if..

On Thu, Nov 11, 2021 at 02:36:52PM -0500, Waiman Long wrote:
static inline bool rwsem_try_write_lock(struct rw_semaphore *sem,
- enum writer_wait_state wstate)
+ struct rwsem_waiter *waiter)
{
long count, new;
+ bool first = rwsem_first_waiter(sem) == waiter;
lockdep_assert_held(&sem->wait_lock);
@@ -546,13 +541,14 @@ static inline bool rwsem_try_write_lock(struct rw_semaphore *sem,
do {
bool has_handoff = !!(count & RWSEM_FLAG_HANDOFF);
- if (has_handoff && wstate == WRITER_NOT_FIRST)
+ if (has_handoff && !first)
return false;
new = count;
if (count & RWSEM_LOCK_MASK) {
- if (has_handoff || (wstate != WRITER_HANDOFF))
+ if (has_handoff || (!waiter->rt_task &&
+ !time_after(jiffies, waiter->timeout)))
return false;
we delete this whole condition, and..
I don't think we can take out this if test.

new |= RWSEM_FLAG_HANDOFF;
@@ -889,6 +888,24 @@ rwsem_spin_on_owner(struct rw_semaphore *sem)
}
#endif
+/*
+ * Common code to handle rwsem flags in out_nolock path with wait_lock held.
+ */
+static inline void rwsem_out_nolock_clear_flags(struct rw_semaphore *sem,
+ struct rwsem_waiter *waiter)
+{
+ long flags = 0;
+
+ list_del(&waiter->list);
+ if (list_empty(&sem->wait_list))
+ flags = RWSEM_FLAG_HANDOFF | RWSEM_FLAG_WAITERS;
+ else if (waiter->handoff_set)
+ flags = RWSEM_FLAG_HANDOFF;
take out this else,

+
+ if (flags)
+ atomic_long_andnot(flags, &sem->count);
+}
We get the inherit thing for free, no?

Once HANDOFF is set, new readers are blocked. And then allow any first
waiter to acquire the lock, who cares if it was the one responsible for
having set the HANDOFF bit.

Yes, we can have the policy of inheriting the HANDOFF bit as long as it is consistent which will be the case here with a common out_nolock function. I can go with that. I just have to document this fact in the comment.

Cheers,
Longman