sched: Fix clock_gettime(CLOCK_[PROCESS/THREAD]_CPUTIME_ID) monotonicity

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 <[email protected]>
Cc: [email protected]
Cc: [email protected]
Cc: Linus Torvalds <[email protected]>
Signed-off-by: Peter Zijlstra <[email protected]>
Link: http://lkml.kernel.org/r/[email protected]
Signed-off-by: Ingo Molnar <[email protected]>

kernel/sched/core.c

@@ -139,6 +139,8 @@ 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);
 }
@@ -2431,7 +2433,12 @@ static u64 do_task_delta_exec(struct task_struct *p, struct rq *rq)
 {
 	u64 ns = 0;
 
-	if (task_current(rq, p)) {
+	/*
+	 * Must be ->curr _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_rq) {
 		update_rq_clock(rq);
 		ns = rq_clock_task(rq) - p->se.exec_start;
 		if ((s64)ns < 0)
@@ -2474,8 +2481,10 @@ unsigned long long task_sched_runtime(struct task_struct *p)
 	 * 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