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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 <umgwanakikbuti@gmail.com> Cc: kosaki.motohiro@jp.fujitsu.com Cc: pjt@google.com Cc: Linus Torvalds <torvalds@linux-foundation.org> Signed-off-by: Peter Zijlstra <peterz@infradead.org> Link: http://lkml.kernel.org/r/1403588980.29711.11.camel@marge.simpson.net Signed-off-by: Ingo Molnar <mingo@kernel.org>
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@ -139,6 +139,8 @@ void update_rq_clock(struct rq *rq)
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return;
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delta = sched_clock_cpu(cpu_of(rq)) - rq->clock;
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if (delta < 0)
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return;
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rq->clock += delta;
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update_rq_clock_task(rq, delta);
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}
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@ -2431,7 +2433,12 @@ static u64 do_task_delta_exec(struct task_struct *p, struct rq *rq)
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{
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u64 ns = 0;
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if (task_current(rq, p)) {
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/*
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* Must be ->curr _and_ ->on_rq. If dequeued, we would
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* project cycles that may never be accounted to this
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* thread, breaking clock_gettime().
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*/
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if (task_current(rq, p) && p->on_rq) {
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update_rq_clock(rq);
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ns = rq_clock_task(rq) - p->se.exec_start;
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if ((s64)ns < 0)
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@ -2474,8 +2481,10 @@ unsigned long long task_sched_runtime(struct task_struct *p)
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* If we race with it leaving cpu, we'll take a lock. So we're correct.
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* If we race with it entering cpu, unaccounted time is 0. This is
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* indistinguishable from the read occurring a few cycles earlier.
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* If we see ->on_cpu without ->on_rq, the task is leaving, and has
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* been accounted, so we're correct here as well.
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*/
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if (!p->on_cpu)
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if (!p->on_cpu || !p->on_rq)
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return p->se.sum_exec_runtime;
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#endif
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