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Date: Tue, 5 Jul 2016 14:04:26 +0200
From: Peter Zijlstra <peterz@infradead.org>
To: Mark Rutland <mark.rutland@arm.com>
Cc: linux-kernel@vger.kernel.org,
        Alexander Shishkin <alexander.shishkin@linux.intel.com>,
        Arnaldo Carvalho de Melo <acme@kernel.org>,
        Ingo Molnar <mingo@redhat.com>, Will Deacon <will.deacon@arm.com>
Subject: Re: [PATCH] perf: fix pmu::filter_match for SW-led groups
Message-ID: <20160705120426.GA30921@twins.programming.kicks-ass.net>
References: <1465917041-15339-1-git-send-email-mark.rutland@arm.com>
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 <20160704180534.GD9048@leverpostej>
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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.