Re: [PATCH v2 05/11] pinctrl: mediatek: avoid virtual gpio trying to set reg
From: Linus Walleij <hidden>
Date: 2019-08-23 08:57:18
Also in:
linux-clk, linux-mediatek, lkml
On Mon, Aug 19, 2019 at 11:22 AM Mars Cheng [off-list ref] wrote:
for virtual gpios, they should not do reg setting and should behave as expected for eint function. Signed-off-by: Mars Cheng <redacted>
This does not explain what a "virtual GPIO" is in this context, so please elaborate. What is this? Why does it exist? What is it used for? GPIO is "general purpose input/output" and it is a pretty rubbery category already as it is, so we need to define our terms pretty strictly.
+bool mtk_is_virt_gpio(struct mtk_pinctrl *hw, unsigned int gpio_n)
+{
+ const struct mtk_pin_desc *desc;
+ bool virt_gpio = false;
+
+ if (gpio_n >= hw->soc->npins)
+ return virt_gpio;
+
+ desc = (const struct mtk_pin_desc *)&hw->soc->pins[gpio_n];
+
+ if (desc->funcs &&
+ desc->funcs[desc->eint.eint_m].name == 0)NULL check is done like this: if (desc->funcs && !desc->funcs[desc->eint.eint_m].name)
+ virt_gpio = true;
So why is this GPIO "virtual" because it does not have a name in the funcs table?
quoted hunk ↗ jump to hunk
@@ -278,6 +295,9 @@ static int mtk_xt_set_gpio_as_eint(void *data, unsigned long eint_n) if (err) return err; + if (mtk_is_virt_gpio(hw, gpio_n)) + return 0;
So does this mean we always succeed in setting a GPIO as eint if it is virtual? Why? Explanatory comment is needed.
quoted hunk ↗ jump to hunk
@@ -693,6 +693,9 @@ static int mtk_gpio_get_direction(struct gpio_chip *chip, unsigned int gpio) const struct mtk_pin_desc *desc; int value, err; + if (mtk_is_virt_gpio(hw, gpio)) + return 1;
Why are "virtual GPIOs" always inputs? Yours, Linus Walleij