Re: [RFC PATCH v2 11/17] sched: Basic tracking of matching tasks

[Aubrey Li]

On Fri, May 3, 2019 at 8:06 AM Tim Chen <tim.c.chen@xxxxxxxxxxxxxxx> wrote:
>
> On 5/1/19 4:27 PM, Tim Chen wrote:
> > On 4/28/19 11:15 PM, Aaron Lu wrote:
> >> On Tue, Apr 23, 2019 at 04:18:16PM +0000, Vineeth Remanan Pillai wrote:
> >>> +/*
> >>> + * Find left-most (aka, highest priority) task matching @cookie.
> >>> + */
> >>> +struct task_struct *sched_core_find(struct rq *rq, unsigned long cookie)
> >>> +{
> >>> +   struct rb_node *node = rq->core_tree.rb_node;
> >>> +   struct task_struct *node_task, *match;
> >>> +
> >>> +   /*
> >>> +    * The idle task always matches any cookie!
> >>> +    */
> >>> +   match = idle_sched_class.pick_task(rq);
> >>> +
> >>> +   while (node) {
> >>> +           node_task = container_of(node, struct task_struct, core_node);
> >>> +
> >>> +           if (node_task->core_cookie < cookie) {
> >>> +                   node = node->rb_left;
> >>
> >> Should go right here?
> >>
> >
> > I think Aaron is correct.  We order the rb tree where tasks with smaller core cookies
> > go to the left part of the tree.
> >
> > In this case, the cookie we are looking for is larger than the current node's cookie.
> > It seems like we should move to the right to look for a node with matching cookie.
> >
> > At least making the following change still allow us to run the system stably for sysbench.
> > Need to gather more data to see how performance changes.
>
> Pawan ran an experiment setting up 2 VMs, with one VM doing a parallel kernel build and one VM doing sysbench,
> limiting both VMs to run on 16 cpu threads (8 physical cores), with 8 vcpu for each VM.
> Making the fix did improve kernel build time by 7%.

I'm gonna agree with the patch below, but just wonder if the testing
result is consistent,
as I didn't see any improvement in my testing environment.

IIUC, from the code behavior, especially for 2 VMs case(only 2
different cookies), the
per-rq rb tree unlikely has nodes with different cookies, that is, all
the nodes on this
tree should have the same cookie, so:
- if the parameter cookie is equal to the rb tree cookie, we meet a
match and go the
third branch
- else, no matter we go left or right, we can't find a match, and
we'll return idle thread
finally.

Please correct me if I was wrong.

Thanks,
-Aubrey
>
> Tim
>
>
> >
> > diff --git a/kernel/sched/core.c b/kernel/sched/core.c
> > index 25638a47c408..ed4cfa49e3f2 100644
> > --- a/kernel/sched/core.c
> > +++ b/kernel/sched/core.c
> > @@ -208,9 +208,9 @@ static struct task_struct *sched_core_find(struct rq *rq, unsigned long cookie)
> >         while (node) {
> >                 node_task = container_of(node, struct task_struct, core_node);
> >
> > -               if (node_task->core_cookie < cookie) {
> > +               if (cookie < node_task->core_cookie) {
> >                         node = node->rb_left;
> > -               } else if (node_task->core_cookie > cookie) {
> > +               } else if (cookie > node_task->core_cookie) {
> >                         node = node->rb_right;
> >                 } else {
> >                         match = node_task;
> >
> >
>