Messages in this thread | | | From | Geert Uytterhoeven <> | Date | Mon, 20 Dec 2021 15:26:44 +0100 | Subject | Re: [PATCH] thermal: rcar_thermal: Use platform_get_irq_optional() to get the interrupt |
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Hi Daniel,
On Mon, Dec 20, 2021 at 3:19 PM Daniel Lezcano <daniel.lezcano@linaro.org> wrote: > On 20/12/2021 14:48, Geert Uytterhoeven wrote: > > On Mon, Dec 20, 2021 at 1:29 PM Daniel Lezcano > > <daniel.lezcano@linaro.org> wrote: > >> On 18/12/2021 15:41, Lad Prabhakar wrote: > >>> platform_get_resource(pdev, IORESOURCE_IRQ, ..) relies on static > >>> allocation of IRQ resources in DT core code, this causes an issue > >>> when using hierarchical interrupt domains using "interrupts" property > >>> in the node as this bypasses the hierarchical setup and messes up the > >>> irq chaining. > >>> > >>> In preparation for removal of static setup of IRQ resource from DT core > >>> code use platform_get_irq_optional(). > >>> > >>> Signed-off-by: Lad Prabhakar <prabhakar.mahadev-lad.rj@bp.renesas.com>
> >>> --- a/drivers/thermal/rcar_thermal.c > >>> +++ b/drivers/thermal/rcar_thermal.c > >>> @@ -445,7 +445,7 @@ static int rcar_thermal_probe(struct platform_device *pdev) > >>> struct rcar_thermal_common *common; > >>> struct rcar_thermal_priv *priv; > >>> struct device *dev = &pdev->dev; > >>> - struct resource *res, *irq; > >>> + struct resource *res; > >>> const struct rcar_thermal_chip *chip = of_device_get_match_data(dev); > >>> int mres = 0; > >>> int i; > >>> @@ -467,9 +467,16 @@ static int rcar_thermal_probe(struct platform_device *pdev) > >>> pm_runtime_get_sync(dev); > >>> > >>> for (i = 0; i < chip->nirqs; i++) { > >>> - irq = platform_get_resource(pdev, IORESOURCE_IRQ, i); > >>> - if (!irq) > >>> + int irq; > >>> + > >>> + irq = platform_get_irq_optional(pdev, i); > >>> + if (irq <= 0 && irq != -ENXIO) { > >>> + ret = irq ? irq : -ENXIO; > >>> + goto error_unregister; > >>> + } > >>> + if (irq == -ENXIO) > >>> continue; > >> > >> Why not invert the conditions? > >> > >> if (irq == -ENXIO) > >> continue; > > > > And this can be break. > > > >> > >> if (irq <= 0) { > >> ret = irq ? irq : -ENXIO; > > > > irq == 0 cannot happen. > > > >> goto out_unregister; > >> } > > Sorry, I don't get the two comments. May be I missed something but it > seems for me the results are the same with the inverted conditions or not. > > if (irq <= 0 && irq != -ENXIO) > goto out; > > if (irq == -ENXIO) > continue; > > Can be changed to: > > if (irq != -ENXIO) > if (irq <= 0) > goto out; > > if (irq == -ENXIO) > continue; > > Can be changed to: > > > if (irq == -ENXIO) > continue; > > if (irq != -ENXIO) > if (irq <= 0) > goto out; > > The second condition is always true because the first condition is the > opposite of the second condition, if the second condition block is > reached, that means irq != -ENXIO, so we can remove the second condition > and that results into: > > if (irq == -ENXIO) > continue; > > if (irq <= 0) > goto out; > > > Did I miss your point ?
I think so, as I don't see your point, neither ;-)
I meant (a) there is no need to continue the loop when there are no more interrupts present, and (b) irq == 0 cannot happen, so the cod can be simplified to:
if (irq == -ENXIO) break; if (irq < 0) { ret = irq; goto out_unregister; }
Gr{oetje,eeting}s,
Geert
-- Geert Uytterhoeven -- There's lots of Linux beyond ia32 -- geert@linux-m68k.org
In personal conversations with technical people, I call myself a hacker. But when I'm talking to journalists I just say "programmer" or something like that. -- Linus Torvalds
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