Jeremy Fitzhardinge wrote:
>> Yeah, disable interrupts, and set a flag that the fake "sti" can test, and
>> just return without doing anything.
>>
>> (You may or may not also need to do extra work to Ack the hardware
>> interrupt etc, which may be irq-controller specific. Once the CPU has
>> accepted the interrupt, you may not be able to just leave it dangling)
>>
>>
>
> So it would be something like:
>
> pda.intr_mask = 1; /* disable interrupts */
> ...
> pda.intr_mask = 0; /* enable interrupts */
> if (xchg(&pda.intr_pending, 0)) /* check pending */
>
Well, can't do xchg, since it implies #LOCK, and you'll lose more than
you gain on the processors where it matters. Cmpxchg is fine, but
processor dependent.
Or, just make the interrupt handlers use software resend for IRQs when
pda.intr_mask is set to zero. Now, local_irq_save / restore are very
pretty:
int local_irq_save(void)
{
int tmp = pda.intr_mask;
pda.intr_mask = 0;
/*
* note there is a window here where local IRQs notice intr_mask == 0
* in that case, they will attempt to resend the IRQ via a tasklet,
* and will succeed, albeit through a slightly longer path
*/
local_bh_disable();
return tmp;
}
void local_irq_restore(int enabled)
{
pda.intr_mask = enabled;
/*
* note there is a window here where softirqs are not processed by
* the interrupt handler, but that is not a problem, since it will
* get done here in the outer enable of any nested pair.
*/
if (enabled)
local_bh_enable();
}
I think Ingo's suggestion of using the hardirq tracing is another way
that could work, but it seems to be too heavyweight and tied too much to
the lockdep verification code - plus it inserts additional
raw_irq_disable in places that seem counter to the goal of getting rid
of them in the first place. Perhaps I'm misunderstanding the code, though.
Zach
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