Re: [PATCH v2 1/4] locking/mutex: Add waiter parameter to mutex_optimistic_spin()

[Peter Zijlstra]

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.