Re: [RFC PATCH] sched/eevdf: Return leftmost entity in pick_eevdf() if no eligible entity is found

Abel Wu
2024-02-29T09:01:40Z

Hi Chen, thanks for detailed analysis.

The title of this patch sounds a little weird to me, since any
non-empty cfs_rq should have at least one eligible entity. Besides,
choosing the leftmost entity which could be non-eligible can be
sub-optimal, anyway this is only a workaround..

On 2/26/24 4:23 PM, Chen Yu Wrote:
> There is occasional report from lkp that the kernel hits the NULL pointer
> exception:
>
> [ 512.079810][ T8305] BUG: kernel NULL pointer dereference, address: 0000002c
> [ 512.080897][ T8305] #PF: supervisor read access in kernel mode
> [ 512.081636][ T8305] #PF: error_code(0x0000) - not-present page
> [ 512.082337][ T8305] *pde = 00000000
> [ 512.082829][ T8305] Oops: 0000 [#1] PREEMPT SMP
> [ 512.083407][ T8305] CPU: 1 PID: 8305 Comm: watchdog Tainted: G W
> [ 512.086203][ T8305] EIP: set_next_entity (fair.c:?)
>
> This is caused by NULL candidate returned by pick_eevdf() as Abel analyzed.
> After
> commit 2227a957e1d5 ("sched/eevdf: Sort the rbtree by virtual deadline")
> the NULL candidate would trigger the NULL pointer exception. While before
> this commit, there would be warning.
>
> This NULL entity issue was always there before above commit. With debug
> patch to print the cfs_rq and all the entities in the tree, we have the
> information when the issue was reproduced:
>
> [ 514.461242][ T8390] cfs_rq avg_vruntime:386638640128 avg_load:2048 min_vruntime:763383370431
> [ 514.535935][ T8390] current on_rq se 0xc5851400, deadline:18435852013562231446
> min_vruntime:18437121115753667698 vruntime:18435852013561943404, load:629
> [ 514.536772][ T8390] Traverse rb-tree from left to right
> [ 514.537138][ T8390] se 0xec1234e0 deadline:763384870431 min_vruntime:763383370431 vruntime:763383370431 non-eligible
> [ 514.537835][ T8390] se 0xec4fcf20 deadline:763762447228 min_vruntime:763760947228 vruntime:763760947228 non-eligible
> [ 514.538539][ T8390] Traverse rb-tree from topdown
> [ 514.538877][ T8390] middle se 0xec1234e0 deadline:763384870431 min_vruntime:763383370431 vruntime:763383370431 non-eligible
> [ 514.539605][ T8390] middle se 0xec4fcf20 deadline:763762447228 min_vruntime:763760947228 vruntime:763760947228 non-eligible
> [ 514.540340][ T8390] Found best:0x0
> [ 514.540613][ T8390] BUG: kernel NULL pointer dereference, address: 00000074
>
> We can see that non of the entities in the tree are eligible, neither is
> the current entity on this cfs_rq. As a result, curr is set to NULL:
> if (curr && (!curr->on_rq || !entity_eligible(cfs_rq, curr)))
> curr = NULL;
>
> and the best is set to NULL, which caused the problem:
> if (!best || (curr && entity_before(curr, best)))
> best = curr;
>
> The cause is that, the curr is eligible, but vruntime_eligible()
> returns false. And the false negative is due to the following
> code in vruntime_eligible():
>
> return avg >= (s64)(vruntime - cfs_rq->min_vruntime) * load;
>
> According to the log, vruntime is 18435852013561943404, the
> cfs_rq->min_vruntime is 763383370431, the load is 629 + 2048 = 2677,
> thus:
> s64 delta = (s64)(18435852013561943404 - 763383370431) = -10892823530978643
> delta * 2677 = 7733399554989275921
> that is to say, the multiply result overflow the s64, which turns the
> negative value into a positive value, thus eligible check fails.

Indeed.

>
> So where is this insane huge vruntime 18435852013561943404 coming from?
> My guess is that, it is because the initial value of cfs_rq->min_vruntime
> is set to (unsigned long)(-(1LL << 20)). If the task(watchdog in this case)
> seldom scheduled in, its vruntime might not move forward too much and
> remain its original value by previous place_entity().

So why not just initialize to 0? The (unsigned long)(-(1LL << 20))
thing is dangerous as it can easily blow up lots of calculations in
lag, key, avg_vruntime and so on.

Say during this pre-life, which is about 1ms for 1024-weight entity,
there is only one entity running in this cfs_rq. Now another entity
with funny lag joins in, being placed somewhere at 0+ vruntime, so
cfs_rq->min_vruntime needs to be adjusted accordingly which leads to
the breakage of cfs_rq->curr's key as you showed above.

>
> The proper fix should deal with the overflow of entity_key() * load, but
> I don't have much clue on that, so propose this conservative method to
> restore the previous behavior before the mentioned commit.

Inspired by Xuewen's proposal, will it work if limit the key?

Thanks & BR,
Abel

>
> Fixes: 2227a957e1d5 ("sched/eevdf: Sort the rbtree by virtual deadline")
> Reported-by: kernel test robot <oliver.sang@intel.com>
> Closes: https://lore.kernel.org/lkml/202401301012.2ed95df0-oliver.sang@intel.com/
> Signed-off-by: Chen Yu <yu.c.chen@intel.com>
> ---
> kernel/sched/fair.c | 13 ++++++++++++-
> 1 file changed, 12 insertions(+), 1 deletion(-)
>
> diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c
> index 533547e3c90a..fb9202f464e2 100644
> --- a/kernel/sched/fair.c
> +++ b/kernel/sched/fair.c
> @@ -880,7 +880,7 @@ static struct sched_entity *pick_eevdf(struct cfs_rq *cfs_rq)
> struct rb_node *node = cfs_rq->tasks_timeline.rb_root.rb_node;
> struct sched_entity *se = __pick_first_entity(cfs_rq);
> struct sched_entity *curr = cfs_rq->curr;
> - struct sched_entity *best = NULL;
> + struct sched_entity *best = NULL, *leftmost;
>
> /*
> * We can safely skip eligibility check if there is only one entity
> @@ -905,6 +905,8 @@ static struct sched_entity *pick_eevdf(struct cfs_rq *cfs_rq)
> goto found;
> }
>
> + leftmost = se;
> +
> /* Heap search for the EEVD entity */
> while (node) {
> struct rb_node *left = node->rb_left;
> @@ -937,6 +939,15 @@ static struct sched_entity *pick_eevdf(struct cfs_rq *cfs_rq)
> if (!best || (curr && entity_before(curr, best)))
> best = curr;
>
> + /*
> + * entity_eligible() could bring false negative due to
> + * multiply overflow, which reports no eligible entity.
> + * Return leftmost entity as a backup(it is guaranteed
> + * the tree is not NULL.
> + */
> + if (!best)
> + best = leftmost;
> +
> return best;
> }
>