Re: [RFC] Deadlock via recursive wakeup via RCU with threadirqs

Sebastian Andrzej Siewior
2019-06-28T18:20:56Z

On 2019-06-28 14:07:27 [-0400], Joel Fernandes wrote:
> On Fri, Jun 28, 2019 at 07:45:45PM +0200, Sebastian Andrzej Siewior wrote:
> > On 2019-06-28 10:30:11 [-0700], Paul E. McKenney wrote:
> > > > I believe the .blocked field remains set even though we are not any more in a
> > > > reader section because of deferred processing of the blocked lists that you
> > > > mentioned yesterday.
> > >
> > > That can indeed happen. However, in current -rcu, that would mean
> > > that .deferred_qs is also set, which (if in_irq()) would prevent
> > > the raise_softirq_irqsoff() from being invoked. Which was why I was
> > > asking the questions about whether in_irq() returns true within threaded
> > > interrupts yesterday. If it does, I need to find if there is some way
> > > of determining whether rcu_read_unlock_special() is being called from
> > > a threaded interrupt in order to suppress the call to raise_softirq()
> > > in that case.
> >
> > Please not that:
> > | void irq_exit(void)
> > | {
> > |…
> > in_irq() returns true
> > | preempt_count_sub(HARDIRQ_OFFSET);
> > in_irq() returns false
> > | if (!in_interrupt() && local_softirq_pending())
> > | invoke_softirq();
> >
> > -> invoke_softirq() does
> > | if (!force_irqthreads) {
> > | __do_softirq();
> > | } else {
> > | wakeup_softirqd();
> > | }
> >
>
> In my traces which I shared previous email, the wakeup_softirqd() gets
> called.
>
> I thought force_irqthreads value is decided at boot time, so I got lost a bit
> with your comment.

It does. I just wanted point out that in this case
rcu_unlock() / rcu_read_unlock_special() won't see in_irq() true.

Sebastian