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sched/numa: Fix a possible divide-by-zero
sched_clock_cpu() may not be consistent between CPUs. If a task migrates to another CPU, then se.exec_start is set to that CPU's rq_clock_task() by update_stats_curr_start(). Specifically, the new value might be before the old value due to clock skew. So then if in numa_get_avg_runtime() the expression: 'now - p->last_task_numa_placement' ends up as -1, then the divider '*period + 1' in task_numa_placement() is 0 and things go bang. Similar to update_curr(), check if time goes backwards to avoid this. [ peterz: Wrote new changelog. ] [ mingo: Tweaked the code comment. ] Signed-off-by: Xie XiuQi <xiexiuqi@huawei.com> Signed-off-by: Peter Zijlstra (Intel) <peterz@infradead.org> Cc: Linus Torvalds <torvalds@linux-foundation.org> Cc: Peter Zijlstra <peterz@infradead.org> Cc: Thomas Gleixner <tglx@linutronix.de> Cc: cj.chengjian@huawei.com Cc: <stable@vger.kernel.org> Link: http://lkml.kernel.org/r/20190425080016.GX11158@hirez.programming.kicks-ass.net Signed-off-by: Ingo Molnar <mingo@kernel.org>
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@ -2007,6 +2007,10 @@ static u64 numa_get_avg_runtime(struct task_struct *p, u64 *period)
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if (p->last_task_numa_placement) {
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delta = runtime - p->last_sum_exec_runtime;
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*period = now - p->last_task_numa_placement;
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/* Avoid time going backwards, prevent potential divide error: */
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if (unlikely((s64)*period < 0))
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*period = 0;
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} else {
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delta = p->se.avg.load_sum;
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*period = LOAD_AVG_MAX;
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