Re: [PATCHv3 2/6] tty/ldsem: Update waiter->task before waking up reader

Peter Zijlstra
2018-09-11T11:45:15Z

On Tue, Sep 11, 2018 at 02:04:49PM +0900, Sergey Senozhatsky wrote:
> On (09/11/18 02:48), Dmitry Safonov wrote:
> > There is a couple of reports about lockup in ldsem_down_read() without
> > anyone holding write end of ldisc semaphore:
> > lkml.kernel.org/r/<20171121132855.ajdv4k6swzhvktl6@wfg-t540p.sh.intel.com>
> > lkml.kernel.org/r/<20180907045041.GF1110@shao2-debian>
> >
> > They all looked like a missed wake up.
> > I wasn't lucky enough to reproduce it, but it seems like reader on
> > another CPU can miss waiter->task update and schedule again, resulting
> > in indefinite (MAX_SCHEDULE_TIMEOUT) sleep.
>
> Certainly, something suspicious is going on.
>
> > @@ -118,6 +118,8 @@ static void __ldsem_wake_readers(struct ld_semaphore *sem)
> > tsk = waiter->task;
> > smp_mb();
> > waiter->task = NULL;
> > + /* Make sure down_read_failed() will see !waiter->task update */
> > + smp_wmb();
> > wake_up_process(tsk);
>
> Hmm. I think wake_up_process() executes a full memory barrier, because
> it accesses task state.
>
> > put_task_struct(tsk);
> > }
> > @@ -217,7 +219,7 @@ down_read_failed(struct ld_semaphore *sem, long count, long timeout)
> > for (;;) {
> > set_current_state(TASK_UNINTERRUPTIBLE);
>
> I think that set_current_state() also executes memory barrier. Just
> because it accesses task state.

In both cases, the rationale, 'because it accesses task state' is
utterly wrong.

The reasoning can be found in the comment near set_current_state().