Re: [RFC -v6 PATCH 4/8] sched: Add yield_to(task, preempt) functionality

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On 01/24/2011 01:12 PM, Peter Zijlstra wrote:
On Thu, 2011-01-20 at 16:34 -0500, Rik van Riel wrote:
From: Mike Galbraith<efault@xxxxxx>

Currently only implemented for fair class tasks.

Add a yield_to_task method() to the fair scheduling class. allowing the
caller of yield_to() to accelerate another thread in it's thread group,
task group.

Implemented via a scheduler hint, using cfs_rq->next to encourage the
target being selected.  We can rely on pick_next_entity to keep things
fair, so noone can accelerate a thread that has already used its fair
share of CPU time.

This also means callers should only call yield_to when they really
mean it.  Calling it too often can result in the scheduler just
ignoring the hint.

Signed-off-by: Rik van Riel<riel@xxxxxxxxxx>
Signed-off-by: Marcelo Tosatti<mtosatti@xxxxxxxxxx>
Signed-off-by: Mike Galbraith<efault@xxxxxx>

Patch 5 wants to be merged back in here I think..

Agreed, but I wanted Mike's comments first  :)


+/**
+ * yield_to - yield the current processor to another thread in
+ * your thread group, or accelerate that thread toward the
+ * processor it's on.
+ *
+ * It's the caller's job to ensure that the target task struct
+ * can't go away on us before we can do any checks.
+ *
+ * Returns true if we indeed boosted the target task.
+ */
+bool __sched yield_to(struct task_struct *p, bool preempt)
+{
+	struct task_struct *curr = current;
+	struct rq *rq, *p_rq;
+	unsigned long flags;
+	bool yielded = 0;
+
+	local_irq_save(flags);
+	rq = this_rq();
+
+again:
+	p_rq = task_rq(p);
+	double_rq_lock(rq, p_rq);
+	while (task_rq(p) != p_rq) {
+		double_rq_unlock(rq, p_rq);
+		goto again;
+	}
+
+	if (!curr->sched_class->yield_to_task)
+		goto out;
+
+	if (curr->sched_class != p->sched_class)
+		goto out;
+
+	if (task_running(p_rq, p) || p->state)
+		goto out;
+
+	if (!same_thread_group(p, curr))
+		goto out;
+
+#ifdef CONFIG_FAIR_GROUP_SCHED
+	if (task_group(p) != task_group(curr))
+		goto out;
+#endif
+
+	yielded = curr->sched_class->yield_to_task(rq, p, preempt);
+
+out:
+	double_rq_unlock(rq, p_rq);
+	local_irq_restore(flags);
+
+	if (yielded)
+		yield();

Calling yield() here is funny, you just had all the locks to actually do
it..

This is us giving up the CPU, which requires not holding locks.

A different thing than us giving the CPU away to someone else.

+static bool yield_to_task_fair(struct rq *rq, struct task_struct *p, bool preempt)
+{
+	struct sched_entity *se =&p->se;
+
+	if (!se->on_rq)
+		return false;
+
+	/* Tell the scheduler that we'd really like pse to run next. */
+	set_next_buddy(se);
+
+	/* Make p's CPU reschedule; pick_next_entity takes care of fairness. */
+	if (preempt)
+		resched_task(rq->curr);
+
+	return true;
+}

So here we set ->next, we could be ->last, and after this we'll set
->yield to curr by calling yield().

So if you do this cyclically I can see ->yield == {->next,->last}
happening.

That would only happen if we called yield_to with ourselves
as the argument!

There is no caller in the tree that does that - task p is
another task, not ourselves.
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