Re: [RFC][PATCH -v3 10/10] locking/mutex: Implement alternative HANDOFF
From: Waiman Long
Date: Fri Sep 09 2016 - 18:32:19 EST
On 09/05/2016 08:36 AM, Peter Zijlstra wrote:
As mentioned in a previous patch, its possible to implement the
handoff logic differently, avoiding the issue where we 'leak' a HADOFF
flag.
This patch does so, just to show what it looks like; I'm not at all
convinced this is worth it.
Signed-off-by: Peter Zijlstra (Intel)<peterz@xxxxxxxxxxxxx>
I have another way of catching the uncleared handoff flag. See the
following code to see if you think that will work.
diff --git a/kernel/locking/mutex.c b/kernel/locking/mutex.c
index 9492494..362ff83 100644
--- a/kernel/locking/mutex.c
+++ b/kernel/locking/mutex.c
@@ -85,7 +85,13 @@ static inline bool __mutex_trylock(struct mutex
*lock, const
owner = atomic_long_read(&lock->owner);
for (;;) { /* must loop, can race against a flag */
- unsigned long old;
+ unsigned long old, flags = __owner_flags(owner);
+
+ /*
+ * We don't need to keep the HANDOFF flag for the waiter.
+ */
+ if (handoff)
+ flags &= ~MUTEX_FLAG_HANDOFF;
if (__owner_task(owner)) {
if (handoff && unlikely(__owner_task(owner) ==
current))
@@ -107,7 +113,7 @@ static inline bool __mutex_trylock(struct mutex
*lock, const
}
old = atomic_long_cmpxchg_acquire(&lock->owner, owner,
- curr |
__owner_flags(owner));
+ curr | flags);
if (old == owner)
return true;
@@ -688,7 +694,7 @@ __mutex_lock_common(struct mutex *lock, long state,
unsigned
* state back to RUNNING and fall through the next
schedule(),
* or we must see its unlock and acquire.
*/
- if (__mutex_trylock(lock, true))
+ if (__mutex_trylock(lock, first))
break;
spin_lock_mutex(&lock->wait_lock, flags);
@@ -700,8 +706,6 @@ remove_waiter:
mutex_remove_waiter(lock, &waiter, task);
if (likely(list_empty(&lock->wait_list)))
__mutex_clear_flag(lock, MUTEX_FLAGS);
- else if (first && (atomic_long_read(&lock->owner) &
MUTEX_FLAG_HANDOFF))
- __mutex_clear_flag(lock, MUTEX_FLAG_HANDOFF);
debug_mutex_free_waiter(&waiter);
Cheers,
Longman