Quoting Maulik Shah (2020-05-27 04:26:14)
> On 5/27/2020 3:14 PM, Stephen Boyd wrote:
> > Quoting Maulik Shah (2020-05-23 10:11:10)
> >> diff --git a/drivers/gpio/gpiolib.c b/drivers/gpio/gpiolib.c
> >> index eaa0e20..3810cd0 100644
> >> --- a/drivers/gpio/gpiolib.c
> >> +++ b/drivers/gpio/gpiolib.c
> >> @@ -2465,32 +2465,37 @@ static void gpiochip_irq_relres(struct irq_data *d)
> >> gpiochip_relres_irq(gc, d->hwirq);
> >> }
> >>
> >> +static void gpiochip_irq_mask(struct irq_data *d)
> >> +{
> >> + struct gpio_chip *gc = irq_data_get_irq_chip_data(d);
> >> +
> >> + if (gc->irq.irq_mask)
> >> + gc->irq.irq_mask(d);
> >> + gpiochip_disable_irq(gc, d->hwirq);
> > How does this work in the lazy case when I want to drive the GPIO? Say I
> > have a GPIO that is also an interrupt. The code would look like
> >
> > struct gpio_desc *gpio = gpiod_get(...)
> > unsigned int girq = gpiod_to_irq(gpio)
> >
> > request_irq(girq, ...);
> >
> > disable_irq(girq);
> > gpiod_direction_output(gpio, 1);
> >
> > In the lazy case genirq wouldn't call the mask function until the first
> > interrupt arrived on the GPIO line. If that never happened then wouldn't
> > we be blocked in gpiod_direction_output() when the test_bit() sees
> > FLAG_USED_AS_IRQ? Or do we need irqs to be released before driving
> > gpios?
>
> The client driver can decide to unlazy disable IRQ with below API...
>
> irq_set_status_flags(girq, IRQ_DISABLE_UNLAZY);
>
> This will immediatly invoke mask function (unlazy disable) from genirq,
> even though irq_disable is not implemented.
>
Sure a consumer can disable the lazy feature, but that shouldn't be
required to make this work. The flag was introduced in commit
e9849777d0e2 ("genirq: Add flag to force mask in
disable_irq[_nosync]()") specifically to help devices that can't disable
the interrupt in their own device avoid a double interrupt.
