Re: [PATCH v4 05/17] irq & spin_lock: Add counted interrupt disabling/enabling

[Shrikanth Hegde]

On 8/5/26 7:50 PM, Boqun Feng wrote:
> On Wed, Aug 05, 2026 at 07:40:18PM +0530, Shrikanth Hegde wrote:
> > Hi Boqun,
> >
> > > Something as below? Going to send it to kernel build bot and see if it
> > > works for all configs.
> > >
> > > ----------------->8
> > > diff --git a/include/linux/interrupt_rc.h b/include/linux/interrupt_rc.h
> > > index b9a7f05ecf42..39f30bc65548 100644
> > > --- a/include/linux/interrupt_rc.h
> > > +++ b/include/linux/interrupt_rc.h
> > > @@ -12,6 +12,7 @@
> > >     */
> > >
> > >    #include <linux/irqflags.h>
> > > +#include <linux/debug_locks.h>
> > >    #include <linux/preempt.h>
> > >    #include <linux/processor.h>
> > >    #include <linux/smp.h>
> > > @@ -63,6 +64,12 @@ static inline void local_interrupt_disable(void)
> > >
> > >           new_count = hardirq_disable_enter();
> > >
> > > +       /* Is hardirq disable count overflow soon? */
> > > +       if (IS_ENABLED(CONFIG_DEBUG_PREEMPT))
> > > +               DEBUG_LOCKS_WARN_ON((new_count & HARDIRQ_DISABLE_MASK) +
> > > +                                   (10 << HARDIRQ_DISABLE_SHIFT) >
> > > +                                   HARDIRQ_DISABLE_MASK);
> > > +
> >
> > This needs a return here right? Else we will see warning for 10 times
> > and then overflow happens and we will call _local_interrupt_disable. No?
>
> DEBUG_LOCKS_WARN_ON() uses debug_locks_off() to avoid this, so we won't
> see it 10 times. The reason not using return here, because we would
> introduce unpaired local_interrupt_disable() if we returned:

Yes, it could be a weird case if the overflow actually happens.
So just warning maybe enough to catch such callers.

Maybe your kunit test can actually help test the behavior with the loop
count.

> 	// hardirq disable count is n
> 	local_interrupt_disable(); // hardirq disable count is n + 1
>
> 	local_interrupt_disable(); <- trigger the warning, if we return
> 				   // hardirq disable count is n + 1

Likely i am missing to understand.

Isn't the count incremented earlier than return?
I.e even if return happens it should be n + 2 right?

> 	local_interrupt_enable(); // hardirq disable count is n
>
> 	local_interrupt_enable(); // hardirq disable count is n - 1
>
> Regards,
> Boqun
>
> > Not sure, if below is any better? (Igore whitespace mangling)
> >
> > if (IS_ENABLED(CONFIG_DEBUG_PREEMPT) &&
> >      DEBUG_LOCKS_WARN_ON((preempt_count() & HARDIRQ_DISABLE_MASK) >=
> > 			 HARDIRQ_DISABLE_MASK - (10 << HARDIRQ_DISABLE_SHIFT)))
> > 	return;
> >
> > >           /* Interrupts can happen here, but it's OK, see __irq_exit_rcu(). */
> > >
> > >           if ((new_count & HARDIRQ_DISABLE_MASK) == HARDIRQ_DISABLE_OFFSET)
> > > @@ -73,6 +80,11 @@ static inline void local_interrupt_enable(void)
> > >    {
> > >           int new_count;
> > >
> > > +       /* Unpaired local_interrupt_enable()? Warn and abort. */
> > > +       if (IS_ENABLED(CONFIG_DEBUG_PREEMPT) &&
> > > +           DEBUG_LOCKS_WARN_ON((preempt_count() & HARDIRQ_DISABLE_MASK) == 0))
> > > +               return;
> > > +
> > >           new_count = hardirq_disable_exit();
> > >
> > >           if ((new_count & HARDIRQ_DISABLE_MASK) == 0) >