Re: [PATCH v3 2/3] pinctrl: bcm: Add STB family pin controller driver

[Andrea della Porta <andrea.porta@suse.com>]

Hi Stanmir,

On 11:19 Tue 19 Aug     , Stanimir Varbanov wrote:
> 
> 
> On 8/19/25 11:14 AM, Andrea della Porta wrote:
> > Hi Stanimir,
> > 
> > On 10:40 Tue 19 Aug     , Stanimir Varbanov wrote:
> >> Hi Andrea,
> >>
> >> On 8/11/25 5:46 PM, Andrea della Porta wrote:
> >>> From: "Ivan T. Ivanov" <iivanov@suse.de>
> >>>
> >>> This driver provide pin muxing and configuration functionality
> >>> for BCM2712 SoC used by RPi5. According to [1] this chip is an
> >>> instance of the one used in Broadcom STB  product line.
> >>>
> >>> [1] https://lore.kernel.org/lkml/f6601f73-cb22-4ba3-88c5-241be8421fc3@broadcom.com/
> >>>
> >>> Cc: Jonathan Bell <jonathan@raspberrypi.com>
> >>> Cc: Phil Elwell <phil@raspberrypi.com>
> >>> Signed-off-by: Ivan T. Ivanov <iivanov@suse.de>
> >>> Reviewed-by: Phil Elwell <phil@raspberrypi.com>
> >>> Signed-off-by: Andrea della Porta <andrea.porta@suse.com>
> >>> ---
> >>>  drivers/pinctrl/bcm/Kconfig           |   13 +
> >>>  drivers/pinctrl/bcm/Makefile          |    1 +
> >>>  drivers/pinctrl/bcm/pinctrl-brcmstb.c | 1197 +++++++++++++++++++++++++
> >>>  3 files changed, 1211 insertions(+)
> >>>  create mode 100644 drivers/pinctrl/bcm/pinctrl-brcmstb.c
> >>>
> >>
> >> <snip>
> >>
> >>> +static int brcmstb_pinctrl_probe(struct platform_device *pdev)
> >>> +{
> >>> +	struct device *dev = &pdev->dev;
> >>> +	struct device_node *np = dev->of_node;
> >>> +	const struct brcmstb_pdata *pdata;
> >>> +	const struct of_device_id *match;
> >>> +	struct brcmstb_pinctrl *pc;
> >>> +	const char **names;
> >>> +	int num_pins, i;
> >>> +
> >>> +	match = of_match_node(brcmstb_pinctrl_match, np);
> >>
> >> The 'match' variable is needless, you can drop it.
> > 
> > you mean something like this?
> > 
> > pdata = of_match_node(brcmstb_pinctrl_match, np)->data;
> > 
> 
> No, I meant:
> 
> pdata = of_device_get_match_data(dev)
> 
> Also as a bonus you could move brcmstb_pinctrl_match[] array after .probe.a

Right. Thanks!

Andrea

> 
> ~Stan