On 02/16/2016 03:53 AM, Peter Zijlstra wrote:
On Mon, Feb 15, 2016 at 06:22:14PM -0800, Jason Low wrote:
On Mon, 2016-02-15 at 18:15 -0800, Jason Low wrote:
On Fri, 2016-02-12 at 14:14 -0800, Davidlohr Bueso wrote:
On Fri, 12 Feb 2016, Peter Zijlstra wrote:
On Fri, Feb 12, 2016 at 12:32:12PM -0500, Waiman Long wrote:
static bool mutex_optimistic_spin(struct mutex *lock,
+ struct ww_acquire_ctx *ww_ctx,
+ const bool use_ww_ctx, int waiter)
{
struct task_struct *task = current;
+ bool acquired = false;
+ if (!waiter) {
+ if (!mutex_can_spin_on_owner(lock))
+ goto done;
Why doesn't the waiter have to check mutex_can_spin_on_owner() ?
afaict because mutex_can_spin_on_owner() fails immediately when the counter
is -1, which is a nono for the waiters case.
mutex_can_spin_on_owner() returns false if the task needs to reschedule
or if the lock owner is not on_cpu. In either case, the task will end up
not spinning when it enters the spin loop. So it makes sense if the
waiter also checks mutex_can_spin_on_owner() so that the optimistic spin
queue overhead can be avoided in those cases.
Actually, since waiters bypass the optimistic spin queue, that means the
the mutex_can_spin_on_owner() isn't really beneficial. So Waiman is
right in that it's fine to skip this in the waiter case.
My concern was the 'pointless' divergence between the code-paths. The
less they diverge the easier it is to understand and review.
If it doesn't hurt, please keep it the same. If it does need to diverge,
include a comment on why.
I will keep the preemption, but will still leave out the
mutex_can_spin_on_owner() check for waiter. I will add a comment to
document that.
Cheers,
Longman
