On Tue, Jul 05, 2016 at 02:04:26PM +0200, Peter Zijlstra wrote:
> On Tue, Jul 05, 2016 at 10:44:48AM +0100, Mark Rutland wrote:
> > My bad; I assumed that for both PMUs we'd start at the root, and thus
> > would need to re-sort in order to get the current CPU's PMU ordered
> > first, much like currently with rotation.
> >
> > I guess I'm having difficulty figuring out the structure of that tree.
> > If we can easily/cheaply find the relevant sub-tree then the above isn't
> > an issue.
>
> struct event {
> struct rb_node node;
> int pmu_id;
> s64 lag;
> ...
> };
>
> bool event_less(struct rb_node *a, struct rb_node *b)
> {
> struct event *left = rb_entry(a, struct event, node);
> struct event *right = rb_entry(b, struct event, node);
>
> if (a->pmu_id < b->pmu_id)
> return true;
>
> if (b->pmu_id > a->pmu_id)
> return false;
>
> /* a->pmu_id == b->pmu_id */
> if (a->lag < b->lag)
> return true;
>
> return false;
> }
>
> Will give you a tree with primary order @pmu_id and secondary order
> @lag.
>
> Which you'd iterate like:
>
> for (event = event_find(pmu_id); event->pmu_id == pmu_id; event =
> event_next(event)) {
> }
>
> And get only the events matching @pmu_id in @lag order.
Cheers! Sorry for being thick; I think I understand now.
I'll have a tinker with the idea.
Thanks,
Mark.
