Re: [bug-report] possible s64 overflow in max_vruntime()

[Roman Kagan]

On Tue, Jan 31, 2023 at 11:21:17AM +0800, Chen Yu wrote:
> On 2023-01-27 at 17:18:56 +0100, Vincent Guittot wrote:
> > On Fri, 27 Jan 2023 at 12:44, Peter Zijlstra <peterz@xxxxxxxxxxxxx> wrote:
> > >
> > > On Thu, Jan 26, 2023 at 07:31:02PM +0100, Roman Kagan wrote:
> > >
> > > > > All that only matters for small sleeps anyway.
> > > > >
> > > > > Something like:
> > > > >
> > > > >         sleep_time = U64_MAX;
> > > > >         if (se->avg.last_update_time)
> > > > >           sleep_time = cfs_rq_clock_pelt(cfs_rq) - se->avg.last_update_time;
> > > >
> > > > Interesting, why not rq_clock_task(rq_of(cfs_rq)) - se->exec_start, as
> > > > others were suggesting?  It appears to better match the notion of sleep
> > > > wall-time, no?
> > >
> > > Should also work I suppose. cfs_rq_clock takes throttling into account,
> > > but that should hopefully also not be *that* long, so either should
> > > work.
> >
> > yes rq_clock_task(rq_of(cfs_rq)) should be fine too
> >
> > Another thing to take into account is the sleeper credit that the
> > waking task deserves so the detection should be done once it has been
> > subtracted from vruntime.
> >
> > Last point, when a nice -20 task runs on a rq, it will take a bit more
> > than 2 seconds for the vruntime to be increased by more than 24ms (the
> > maximum credit that a waking task can get) so threshold must be
> > significantly higher than 2 sec. On the opposite side, the lowest
> > possible weight of a cfs rq is 2 which means that the problem appears
> > for a sleep longer or equal to 2^54 = 2^63*2/1024. We should use this
> > value instead of an arbitrary 200 days
> Does it mean any threshold between 2 sec and 2^54 nsec should be fine? Because
> 1. Any task sleeps longer than 2 sec will get at most 24 ms(sysctl_sched_latency)
>    'vruntime bonus' when enqueued.
> 2. Although a low weight cfs rq run for 2^54 nsec could trigger the overflow,
>    we can choose threshold lower than 2^54 to avoid any overflow.

This matches my understanding too, so I went ahead with the value
proposed by Peter (1 min) which looked sufficiently far away from either
side.

Roman.



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