[PATCH 4/9] pinctrl: meson: allow gpio to request irq

Jerome Brunet jbrunet at baylibre.com
Wed Oct 26 07:22:46 PDT 2016


On Tue, 2016-10-25 at 20:20 +0200, Linus Walleij wrote:
> On Tue, Oct 25, 2016 at 5:31 PM, Jerome Brunet <jbrunet at baylibre.com>
> wrote:
> > 
> > On Tue, 2016-10-25 at 15:47 +0100, Marc Zyngier wrote:
> 
> > 
> > > 
> > > Is gpio_to_irq() supposed to allocate an interrupt? Or merely to
> > > report the existence of a mapping?
> 
> It should provide an IRQ corresponding to the gpio line, if possible.
> 
> However the semantic is such, that it is not necessary to call
> to_irq()
> before using an IRQ: the irqchip and gpiochip abstractions should be
> orthogonal.

Linus,

They are orthogonal. You can request an irq from the irqchip controller
without the gpiochip, like any other irq controller.

> 
> This goes especially when using device tree or ACPI, where you
> may reference an IRQ from something modeled as irqchip, which
> is simultaneously a gpiochip.
> 
> > 
> > Linus, please correct me if I'm wrong,
> > .to_irq gets the linux gpio number and returns the linux virtual
> > irq
> > numbers, 0 if there is no interrupt.
> 
> Yes. But it may *or may not* be called before using the IRQ.
> 
> So it should look up or try to create a mapping on request, but not
> assume to have been called before using some line as IRQ.
> 
> The only thing you should assume to be called before an interrupt
> is put to use is the stuff in irqchip. So you have to do your
> dynamic irqdomain mapping elsewhere than .to_irq().

irq_create_mapping (and irq_create_fwspec_mapping) internally calls
irq_find_mapping. So if the mapping already exist (the irq is already
used before calling to_irq), the existing mapping will be returned. The
mapping will be actually created only if needed. It seems to be in line
with your explanation, no ?

There is really a *lot* of gpio drivers which use irq_create_mapping in
the to_irq callback, are these all wrong ?
If this should not be used, what should we all do instead ? 

> 
> Yours,
> Linus Walleij



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