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

diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c
index a4d9e14bf13891482bb717d03946699fa04958b1..35f3ea3750844c2a8789e7b6312bf8432caa1e5a 100644 (file)
--- a/kernel/sched/fair.c
+++ b/kernel/sched/fair.c
@@ -2007,6 +2007,10 @@ static u64 numa_get_avg_runtime(struct task_struct *p, u64 *period)
        if (p->last_task_numa_placement) {
                delta = runtime - p->last_sum_exec_runtime;
                *period = now - p->last_task_numa_placement;
+
+               /* Avoid time going backwards, prevent potential divide error: */
+               if (unlikely((s64)*period < 0))
+                       *period = 0;
        } else {
                delta = p->se.avg.load_sum;
                *period = LOAD_AVG_MAX;