oom: filter tasks not sharing the same cpuset

Tasks that do not share the same set of allowed nodes with the task that
triggered the oom should not be considered as candidates for oom kill.

Tasks in other cpusets with a disjoint set of mems would be unfairly
penalized otherwise because of oom conditions elsewhere; an extreme
example could unfairly kill all other applications on the system if a
single task in a user's cpuset sets itself to OOM_DISABLE and then uses
more memory than allowed.

Killing tasks outside of current's cpuset rarely would free memory for
current anyway.  To use a sane heuristic, we must ensure that killing a
task would likely free memory for current and avoid needlessly killing
others at all costs just because their potential memory freeing is
unknown.  It is better to kill current than another task needlessly.

Signed-off-by: David Rientjes <[email protected]>
Acked-by: Rik van Riel <[email protected]>
Acked-by: Nick Piggin <[email protected]>
Acked-by: Balbir Singh <[email protected]>
Cc: KOSAKI Motohiro <[email protected]>
Reviewed-by: KAMEZAWA Hiroyuki <[email protected]>
Signed-off-by: Andrew Morton <[email protected]>
Signed-off-by: Linus Torvalds <[email protected]>

diff --git a/mm/oom_kill.c b/mm/oom_kill.c
index 0c7c18f784251ed0dc0066534c78a7382bb36846..6f6e04c40c9347b5f1b6556bb0a9a4a08cc4af3a 100644 (file)
--- a/mm/oom_kill.c
+++ b/mm/oom_kill.c
@@ -183,14 +183,6 @@ unsigned long badness(struct task_struct *p, unsigned long uptime)
        if (has_capability_noaudit(p, CAP_SYS_RAWIO))
                points /= 4;
 
-       /*
-        * If p's nodes don't overlap ours, it may still help to kill p
-        * because p may have allocated or otherwise mapped memory on
-        * this node before. However it will be less likely.
-        */
-       if (!has_intersects_mems_allowed(p))
-               points /= 8;
-
        /*
         * Adjust the score by oom_adj.
         */
@@ -277,6 +269,8 @@ static struct task_struct *select_bad_process(unsigned long *ppoints,
                        continue;
                if (mem && !task_in_mem_cgroup(p, mem))
                        continue;
+               if (!has_intersects_mems_allowed(p))
+                       continue;
 
                /*
                 * This task already has access to memory reserves and is