linux-kernel - Re: [PATCH V2 1/3] watchdog: imgpdc: Allow timeout to be set in device-tree

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Date:	Thu, 02 Apr 2015 18:52:58 -0700
From:	Guenter Roeck <linux@...ck-us.net>
To:	Andrew Bresticker <abrestic@...omium.org>
CC:	James Hogan <james.hogan@...tec.com>,
	Wim Van Sebroeck <wim@...ana.be>,
	"linux-watchdog@...r.kernel.org" <linux-watchdog@...r.kernel.org>,
	"linux-kernel@...r.kernel.org" <linux-kernel@...r.kernel.org>,
	Ezequiel Garcia <ezequiel.garcia@...tec.com>
Subject: Re: [PATCH V2 1/3] watchdog: imgpdc: Allow timeout to be set in device-tree

On 04/02/2015 09:46 AM, Andrew Bresticker wrote:
> On Wed, Apr 1, 2015 at 6:08 PM, Guenter Roeck <linux@...ck-us.net> wrote:
>> On 04/01/2015 03:22 PM, James Hogan wrote:
>>>
>>> Hi Andrew,
>>>
>>> On Wed, Apr 01, 2015 at 10:43:14AM -0700, Andrew Bresticker wrote:
>>>>
>>>> Since the heartbeat is statically initialized to its default value,
>>>> watchdog_init_timeout() will never look in the device-tree for a
>>>> timeout-sec value.  Instead of statically initializing heartbeat,
>>>> fall back to the default timeout value if watchdog_init_timeout()
>>>> fails.
>>>
>>>
>>> Whoops. Sorry about that. I wasn't aware that a timeout-sec value was
>>> expected. It isn't mentioned in the DT binding documentation for this
>>> device :-(.
>>>
>>>>
>>>> Signed-off-by: Andrew Bresticker <abrestic@...omium.org>
>>>> Cc: Ezequiel Garcia <ezequiel.garcia@...tec.com>
>>>> Cc: James Hogan <james.hogan@...tec.com>
>>>> ---
>>>> New for v2.
>>>> ---
>>>>    drivers/watchdog/imgpdc_wdt.c | 6 +++---
>>>>    1 file changed, 3 insertions(+), 3 deletions(-)
>>>>
>>>> diff --git a/drivers/watchdog/imgpdc_wdt.c
>>>> b/drivers/watchdog/imgpdc_wdt.c
>>>> index 0deaa4f..89b2abc 100644
>>>> --- a/drivers/watchdog/imgpdc_wdt.c
>>>> +++ b/drivers/watchdog/imgpdc_wdt.c
>>>> @@ -42,7 +42,7 @@
>>>>    #define PDC_WDT_MIN_TIMEOUT           1
>>>>    #define PDC_WDT_DEF_TIMEOUT           64
>>>>
>>>> -static int heartbeat = PDC_WDT_DEF_TIMEOUT;
>>>> +static int heartbeat;
>>>>    module_param(heartbeat, int, 0);
>>>>    MODULE_PARM_DESC(heartbeat, "Watchdog heartbeats in seconds "
>>>>          "(default=" __MODULE_STRING(PDC_WDT_DEF_TIMEOUT) ")");
>>>> @@ -195,9 +195,9 @@ static int pdc_wdt_probe(struct platform_device
>>>> *pdev)
>>>>
>>>>          ret = watchdog_init_timeout(&pdc_wdt->wdt_dev, heartbeat,
>>>> &pdev->dev);
>>>>          if (ret < 0) {
>>>> -               pdc_wdt->wdt_dev.timeout = pdc_wdt->wdt_dev.max_timeout;
>>>> +               pdc_wdt->wdt_dev.timeout = PDC_WDT_DEF_TIMEOUT;
>>>
>>>
>>> The watchdog_init_timeout kerneldoc comment suggests that the old value
>>> should be the default timeout, i.e. that timeout should be set to
>>> PDC_WDT_DEF_TIMEOUT before calling watchdog_init_timeout, rather than
>>> whenever ret < 0.
>>>
>>> Indeed, if heartbeat is set to an invalid non-zero value,
>>> watchdog_init_timeout will still try and set timeout from DT, but also
>>> still returns -EINVAL regardless of whether that succeeds, and this
>>> would incorrectly override the timeout from DT with the hardcoded
>>> default.
>>>
>>>>                  dev_warn(&pdev->dev,
>>>> -                        "Initial timeout out of range! setting max
>>>> timeout\n");
>>>> +                        "Initial timeout out of range! setting default
>>>> timeout\n");
>>>
>>>
>>> It feels wrong for a presumably safe & normal situation (i.e. no default
>>> in DT, which arguably shouldn't contain policy anyway) to show a
>>> warning, but it can also show due to an invalid module parameter (or
>>> invalid DT property) which is most definitely justified.
>>>
>>
>> Agreed. I would suggest to leave that part alone and set the default prior
>> to calling watchdog_init_timeout().
>
> Yes, but I think James' concern here was that we'd now get a
> dev_warn() in the normal case where no timeout is specified via module
> parameter or DT.
>
My understanding is that watchdog_init_timeout only returns an error if
the second parameter is not 0 and invalid, or if the timeout-sec property
has been provided and is invalid. I am not entirely sure I understand
why you think this is a problem. Can you please explain ?

Thanks,
Guenter

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