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.
