Re: [PATCH v3] pinctrl: ocelot: Fix interrupt controller

[Michael Walle <michael@walle.cc>]

Hi Horatiu,

Am 2022-10-07 11:49, schrieb Horatiu Vultur:
> The 10/06/2022 13:43, Michael Walle wrote:
> 
> Hi Walle,
> 
>> Seeing 20 was definitely fishy, seeing two instead of one maybe not
>> so much. I guess it will create one spurious interrupt if none of
>> the registered handlers will care.
>> 
>> OTOH, the code below won't work in all cases anyway, right? It's just
>> best effort.
> 
> I was expecting to work in all cases, but if you found some cases that
> would not work, please point them out.
> 
>> 
>> > Below I have a diff that I tried with LAN8814 PHYs and I could see that
>> > count in /proc/interrupts is increasing correctly.
>> >
>> > > I've verified that there is only one low pulse on the interrupt line.
>> > > I've
>> > > noticed though, that the number of interrupts seem to be correlating
>> > > with
>> > > the length of the low pulse.
>> > ---
>> > diff --git a/drivers/pinctrl/pinctrl-ocelot.c
>> > b/drivers/pinctrl/pinctrl-ocelot.c
>> > index c7df8c5fe5854..105771ff82e62 100644
>> > --- a/drivers/pinctrl/pinctrl-ocelot.c
>> > +++ b/drivers/pinctrl/pinctrl-ocelot.c
>> > @@ -1863,19 +1863,28 @@ static void ocelot_irq_unmask_level(struct
>> > irq_data *data)
>> >       if (val & bit)
>> >               ack = true;
>> >
>> > +     /* Try to clear any rising edges */
>> > +     if (!active && ack)
>> > +             regmap_write_bits(info->map, REG(OCELOT_GPIO_INTR, info, gpio),
>> > +                               bit, bit);
>> 
>> Might we lose interrupts here, if the line would go active again right
>> after the read of the line state and before reading the "ack" bit?
> 
> We lose the interrupt here, as the HW will not generate another one
> but at later point we read again the line status. And if the line is
> active then we kick again the interrupt handler again.

Ahh, thanks for explaining. That also explains the read below.

Will you send a proper patch?

-michael

> 
>> 
>> > +
>> >       /* Enable the interrupt now */
>> >       gpiochip_enable_irq(chip, gpio);
>> >       regmap_update_bits(info->map, REG(OCELOT_GPIO_INTR_ENA, info, gpio),
>> >                          bit, bit);
>> >
>> >       /*
>> > -      * In case the interrupt line is still active and the interrupt
>> > -      * controller has not seen any changes in the interrupt line, then it
>> > -      * means that there happen another interrupt while the line was
>> > active.
>> > +      * In case the interrupt line is still active then it means that
>> > +      * there happen another interrupt while the line was active.
>> >        * So we missed that one, so we need to kick the interrupt again
>> >        * handler.
>> >        */
>> > -     if (active && !ack) {
>> > +     regmap_read(info->map, REG(OCELOT_GPIO_IN, info, gpio), &val);
>> > +     if ((!(val & bit) && trigger_level == IRQ_TYPE_LEVEL_LOW) ||
>> > +           (val & bit && trigger_level == IRQ_TYPE_LEVEL_HIGH))
>> > +             active = true;
>> 
>> Why do you read the line state twice? What happens if the line state
>> changes right after you've read it?
> 
> Here we need to read again the status because we might have clear the
> ack of interrupt.
> If the line becomes active right after this read, then the HW will
> generate another interrupt as the interrupt is enabled and ack is
> cleared.
> 
>> 
>> > +
>> > +     if (active) {
>> >               struct ocelot_irq_work *work;
>> >
>> >               work = kmalloc(sizeof(*work), GFP_ATOMIC);
>> 
>> So yes, maybe the trade-off that there will be two interrupts are
>> better than this additional patch. But it should be documented
>> somewhere, even if it's just a comment in this driver.
>> 
>> -michael