If a task has been dequeued, it has been accounted. Do not project
cycles that may or may not ever be accounted to a dequeued task, as
that may make clock_gettime() both inaccurate and non-monotonic.
Protect update_rq_clock() from slight TSC skew while at it.
Signed-off-by: Mike Galbraith <umgwanakikb...@gmail.com>
---
kernel/sched/core.c | 13 +++++++++++--
1 file changed, 11 insertions(+), 2 deletions(-)
--- a/kernel/sched/core.c
+++ b/kernel/sched/core.c
@@ -144,6 +144,8 @@ void update_rq_clock(struct rq *rq)
return;
delta = sched_clock_cpu(cpu_of(rq)) - rq->clock;
+ if (delta < 0)
+ return;
rq->clock += delta;
update_rq_clock_task(rq, delta);
}
@@ -2533,7 +2535,12 @@ static u64 do_task_delta_exec(struct tas
{
u64 ns = 0;
- if (task_current(rq, p)) {
+ /*
+ * Must be ->curr, ->on_cpu _and_ ->on_rq. If dequeued, we
+ * would project cycles that may never be accounted to this
+ * thread, breaking clock_gettime().
+ */
+ if (task_current(rq, p) && p->on_cpu && p->on_rq) {
update_rq_clock(rq);
ns = rq_clock_task(rq) - p->se.exec_start;
if ((s64)ns < 0)
@@ -2576,8 +2583,10 @@ unsigned long long task_sched_runtime(st
* If we race with it leaving cpu, we'll take a lock. So we're correct.
* If we race with it entering cpu, unaccounted time is 0. This is
* indistinguishable from the read occurring a few cycles earlier.
+ * If we see ->on_cpu without ->on_rq, the task is leaving, and has
+ * been accounted, so we're correct here as well.
*/
- if (!p->on_cpu)
+ if (!p->on_cpu || !p->on_rq)
return p->se.sum_exec_runtime;
#endif
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