Re: [PATCH RT] push RT tasks when overloaded (via schedule)

From: Steven Rostedt
Date: Fri Oct 12 2007 - 13:01:22 EST




--
On Fri, 12 Oct 2007, Peter Zijlstra wrote:

>
> On Fri, 2007-10-12 at 12:05 -0400, Steven Rostedt wrote:
>
> > Index: linux-2.6.23-rt1/kernel/sched.c
> > ===================================================================
> > --- linux-2.6.23-rt1.orig/kernel/sched.c
> > +++ linux-2.6.23-rt1/kernel/sched.c
> > @@ -304,6 +304,7 @@ struct rq {
> > #ifdef CONFIG_PREEMPT_RT
> > unsigned long rt_nr_running;
> > unsigned long rt_nr_uninterruptible;
> > + int curr_prio;
>
> still not convinced we want this PREEMPT_RT only.

Neither am I, but I want testing first ;-)

>
> > #endif
> >
> > unsigned long switch_timestamp;
> > @@ -1484,6 +1485,126 @@ next_in_queue:
> >
> > static int double_lock_balance(struct rq *this_rq, struct rq *busiest);
> >
> > +/* Only try this algorithm three times */
> > +#define RT_PUSH_MAX_TRIES 3
> > +
> > +/* Will lock the rq it finds */
> > +static int find_lowest_cpu(cpumask_t *cpu_mask, struct task_struct *task,
> > + struct rq *this_rq)
> > +{
>
> Eeew, asymetric locking. At the very least name it:
> find_lock_lowest_rq() or something like that.

Hehe, I originally had it that name (well without the lock). But I need
both the found cpu and the rq. So instead of passing in a pointer to the
dst_cpu I just returned it, since it was simple to do a cpu_rq(). But if
you find this too grotesque, I can just return the cpu via a pointer.


>
> Might as well return struct rq* while we're at it [1].

That's what I originally did.

>
> > + struct rq *lowest_rq = NULL;
> > + int dst_cpu = -1;
> > + int cpu;
> > + int tries;
> > +
> > + for (tries = 0; tries < RT_PUSH_MAX_TRIES; tries++) {
> > + /*
> > + * Scan each rq for the lowest prio.
> > + */
> > + for_each_cpu_mask(cpu, *cpu_mask) {
> > + struct rq *rq = &per_cpu(runqueues, cpu);
> > +
> > + if (cpu == smp_processor_id())
> > + continue;
> > +
> > + /* We look for lowest RT prio or non-rt CPU */
> > + if (rq->curr_prio >= MAX_RT_PRIO) {
> > + lowest_rq = rq;
> > + dst_cpu = cpu;
> > + break;
> > + }
> > +
> > + /* no locking for now */
> > + if (rq->curr_prio > task->prio &&
> > + (!lowest_rq || rq->curr_prio < lowest_rq->curr_prio)) {
> > + dst_cpu = cpu;
> > + lowest_rq = rq;
> > + }
> > + }
> > +
> > + if (!lowest_rq) {
> > + dst_cpu = -1;
> > + break;
> > + }
> > +
> > + /* if the prio of this runqueue changed, try again */
> > + if (double_lock_balance(this_rq, lowest_rq)) {
> > + /*
> > + * We had to unlock the run queue. In
> > + * the mean time, task could have
> > + * migrated already or had its affinity changed.
> > + */
> > + if (unlikely(task_rq(task) != this_rq ||
> > + !cpu_isset(dst_cpu, task->cpus_allowed))) {
> > + spin_unlock(&lowest_rq->lock);
> > + dst_cpu = -1;
> > + lowest_rq = NULL;
> > + break;
> > + }
> > +
> > + }
> > +
> > + /* If this rq is still suitable use it. */
> > + if (lowest_rq->curr_prio > task->prio)
> > + break;
> > +
> > + /* try again */
> > + spin_unlock(&lowest_rq->lock);
> > + lowest_rq = NULL;
> > + dst_cpu = -1;
> > + }
> > +
> > + return dst_cpu;
> > +}
> > +
> > +/*
> > + * If the current CPU has more than one RT task, see if the non
> > + * running task can migrate over to a CPU that is running a task
> > + * of lesser priority.
> > + */
> > +static int push_rt_task(struct rq *this_rq)
> > +{
> > + struct task_struct *next_task;
> > + struct rq *lowest_rq;
> > + int dst_cpu;
> > + int ret = 0;
> > + cpumask_t cpu_mask;
> > +
> > + assert_spin_locked(&this_rq->lock);
> > +
> > + next_task = rt_next_highest_task(this_rq);
> > + if (!next_task)
> > + return 0;
> > +
> > + cpus_and(cpu_mask, cpu_online_map, next_task->cpus_allowed);
> > +
> > + /* We might release this_rq lock */
> > + get_task_struct(next_task);
> > +
> > + /* find_lowest_rq locks cpu_rq(dst_cpu) if found */
> > + dst_cpu = find_lowest_cpu(&cpu_mask, next_task, this_rq);
> > + if (dst_cpu < 0)
> > + goto out;
> > +
> > + lowest_rq = cpu_rq(dst_cpu);
> > +
>
> [1] Because that is what we use anyway.
>
> > + assert_spin_locked(&lowest_rq->lock);
> > +
> > + deactivate_task(this_rq, next_task, 0);
> > + set_task_cpu(next_task, dst_cpu);

And here is where i need the dst_cpu. Do you prefer that I pass in the
cpu as a pointer? or just simply use lowest_rq->cpu. I guess that would
work too.

> > + activate_task(lowest_rq, next_task, 0);
> > +
> > + resched_task(lowest_rq->curr);
> > +
> > + spin_unlock(&lowest_rq->lock);
> > +
> > + ret = 1;
> > +out:
> > + put_task_struct(next_task);
> > +
> > + return ret;
> > +}
> > +
> > /*
> > * Pull RT tasks from other CPUs in the RT-overload
> > * case. Interrupts are disabled, local rq is locked.
> > @@ -2202,19 +2323,28 @@ static inline void finish_task_switch(st
> > * be dropped twice.
> > * Manfred Spraul <manfred@xxxxxxxxxxxxxxxx>
> > */
> > + prev_state = prev->state;
> > + _finish_arch_switch(prev);
> > +#if defined(CONFIG_PREEMPT_RT) && defined(CONFIG_SMP)
> > + rq->curr_prio = current->prio;
> > +#endif
> > + finish_lock_switch(rq, prev);
> > #if defined(CONFIG_PREEMPT_RT) && defined(CONFIG_SMP)
> > /*
> > * If we pushed an RT task off the runqueue,
> > - * then kick other CPUs, they might run it:
> > - */
> > - if (unlikely(rt_task(current) && rq->rt_nr_running > 1)) {
> > - schedstat_inc(rq, rto_schedule);
> > - smp_send_reschedule_allbutself_cpumask(current->cpus_allowed);
> > + * then kick other CPUs, they might run it.
> > + * Note we may release the rq lock, and since
> > + * the lock was owned by prev, we need to release it
> > + * first via finish_lock_switch and then reaquire it.
> > + */
> > + if (unlikely(rt_task(current))) {
> > + spin_lock(&rq->lock);
> > + /* push_rt_task will return true if it moved an RT */
> > + while (push_rt_task(rq))
> > + ;
> > + spin_unlock(&rq->lock);
> > }
> > #endif
> > - prev_state = prev->state;
> > - _finish_arch_switch(prev);
> > - finish_lock_switch(rq, prev);
>
> finish_lock_switch() seems to do spin_unlock_irq() in the tree I'm
> looking at. which mean you just 'forgot' to re-enable IRQs.

In my tree (2.6.23-rt1) it is just a spin_unlock().

-- Steve
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