On Thu, Apr 25, 2019 at 11:52:28AM +0800, Xie XiuQi wrote: > On 2019/4/24 2:44, Peter Zijlstra wrote:
> > I'll try and come up with a better Changelog tomorrow. I actually did, but forgot to send out. I have the below. Does that work for you? --- Subject: sched/numa: Fix a possible divide-by-zero From: Xie XiuQi <[email protected]> Date: Sat, 20 Apr 2019 16:34:16 +0800 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. Cc: [email protected] Cc: [email protected] Signed-off-by: Xie XiuQi <[email protected]> [peterz: simplified changelog] Signed-off-by: Peter Zijlstra (Intel) <[email protected]> Link: https://lkml.kernel.org/r/[email protected] --- kernel/sched/fair.c | 4 ++++ 1 file changed, 4 insertions(+) --- a/kernel/sched/fair.c +++ b/kernel/sched/fair.c @@ -2007,6 +2007,10 @@ static u64 numa_get_avg_runtime(struct t if (p->last_task_numa_placement) { delta = runtime - p->last_sum_exec_runtime; *period = now - p->last_task_numa_placement; + + /* Avoid backward, and prevent potential divide error */ + if (unlikely((s64)*period < 0)) + *period = 0; } else { delta = p->se.avg.load_sum; *period = LOAD_AVG_MAX;
