Re: Re: [linus:master] [sched/eevdf] 2227a957e1: BUG:kernel_NULL_pointer_dereference,address

[Chen Yu <yu.c.chen@intel.com>]

On 2024-02-20 at 12:04:33 +0800, Abel Wu wrote:
> On 2/19/24 8:49 PM, Chen Yu Wrote:
> > 
> > While looking at pick_eevdf(), I have a thought.
> > Currently the sched entity is sorted by their deadline. During task
> > pickup, the pick_eevdf() scans for an candidate sched entity with the
> > smallest deadline. Meanwhile this candidate sched entity must also be
> > eligible.
> > 
> > The scan is O(lgn) on average, and O(1) at best case. How about making the
> > average scan even faster by sorting the sched entity not only by deadline,
> > but also the eligibility? The idea is that, the eligible sched entity with
> > smaller deadline is sorted at the front the tree. Otherwise, if the entity
> > is not eligible, even if it has a smaller deadline, it should be sorted
> > at the end of the tree.
> 
> Eligibility is dynamic due to the nature of weighted average vruntime.
> IIUC if doing so like above, update_curr() should take the responsibility
> to re-sort the tree which seems to be O(logN).
> 
> > 
> > After the change, pick_eevdf() get the leftmost sched entity at O(1) on
> > average. Besides, it is guaranteed to return non-NULL sched entity in
> > pick_eevdf(), which prevents suspicious NULL pointer exception in pick_eevdf().
> 
> It is guaranteed when doing pick that the rbtree is non-NULL, and given
> that rq lock is held, I don't think the bug is inside pick_eevdf().
>

That's true, my suspect is that although the tree is not NULL, the eligible
check might return false negative thus no candidate is chosen, and pick_eevdf()
is the victim of that. Previously commit 8dafa9d0eb1a ("sched/eevdf: Fix min_deadline
heap integrity") did similar thing to fix the min_deadline. Without this change,
the tree might be traversed incorrectly and found a NULL candidate. And this is
why I'm thinking of printing the whole tree when NULL entity is chosen in
pick_eevdf().

thanks,
Chenyu