linux-kernel - Re: [PATCH] watchdog: s3c2410_wdt: Handle rounding a little better for timeout

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Date:	Tue, 26 Nov 2013 16:10:24 -0800
From:	Guenter Roeck <linux@...ck-us.net>
To:	Doug Anderson <dianders@...omium.org>
CC:	Wim Van Sebroeck <wim@...ana.be>,
	Leela Krishna Amudala <l.krishna@...sung.com>,
	Olof Johansson <olof@...om.net>,
	Tomasz Figa <tomasz.figa@...il.com>,
	Kukjin Kim <kgene.kim@...sung.com>,
	Ben Dooks <ben-linux@...ff.org>,
	"linux-arm-kernel@...ts.infradead.org" 
	<linux-arm-kernel@...ts.infradead.org>,
	linux-samsung-soc <linux-samsung-soc@...r.kernel.org>,
	linux-watchdog@...r.kernel.org,
	"linux-kernel@...r.kernel.org" <linux-kernel@...r.kernel.org>
Subject: Re: [PATCH] watchdog: s3c2410_wdt: Handle rounding a little better
 for timeout

On 11/26/2013 01:34 PM, Doug Anderson wrote:
> Guenter,
>
> On Tue, Nov 26, 2013 at 10:48 AM, Guenter Roeck <linux@...ck-us.net> wrote:
>> On 11/26/2013 10:30 AM, Doug Anderson wrote:
>>>
>>> The existing watchdog timeout worked OK but didn't deal with
>>> rounding in an ideal way when dividing out all of its clocks.
>>>
>>> Specifically if you had a timeout of 32 seconds and an input clock of
>>> 66666666, you'd end up setting a timeout of 31.9998 seconds and
>>> reporting a timeout of 31 seconds.
>>>
>>> Specifically DBG printouts showed:
>>>     s3c2410wdt_set_heartbeat: count=16666656, timeout=32, freq=520833
>>>     s3c2410wdt_set_heartbeat: timeout=32, divisor=255, count=16666656
>>> (0000ff4f)
>>> and the final timeout reported to the user was:
>>>     ((count / divisor) * divisor) / freq
>>>     (0xff4f * 255) / 520833 = 31 (truncated from 31.9998)
>>> the technically "correct" value is:
>>>     (0xff4f * 255) / (66666666.0 / 128) = 31.9998
>>>
>>> By using "DIV_ROUND_UP" we can be a little more correct.
>>>     s3c2410wdt_set_heartbeat: count=16666688, timeout=32, freq=520834
>>>     s3c2410wdt_set_heartbeat: timeout=32, divisor=255, count=16666688
>>> (0000ff50)
>>> and the final timeout reported to the user:
>>>     (0xff50 * 255) / 520834 = 32
>>> the technically "correct" value is:
>>>     (0xff50 * 255) / (66666666.0 / 128) = 32.0003
>>>
>>> We'll use a DIV_ROUND_UP to solve this, generally erroring on the side
>>> of reporting shorter values to the user and setting the watchdog to
>>> slightly longer than requested:
>>> * Round input frequency up to assume watchdog is counting faster.
>>> * Round divisions by divisor up to give us extra time.
>>>
>>> Signed-off-by: Doug Anderson <dianders@...omium.org>
>>> ---
>>>    drivers/watchdog/s3c2410_wdt.c | 10 +++++-----
>>>    1 file changed, 5 insertions(+), 5 deletions(-)
>>>
>>> diff --git a/drivers/watchdog/s3c2410_wdt.c
>>> b/drivers/watchdog/s3c2410_wdt.c
>>> index 7d8fd04..fe2322b 100644
>>> --- a/drivers/watchdog/s3c2410_wdt.c
>>> +++ b/drivers/watchdog/s3c2410_wdt.c
>>> @@ -188,7 +188,7 @@ static int s3c2410wdt_set_heartbeat(struct
>>> watchdog_device *wdd, unsigned timeou
>>>          if (timeout < 1)
>>>                  return -EINVAL;
>>>
>>> -       freq /= 128;
>>> +       freq = DIV_ROUND_UP(freq, 128);
>>>          count = timeout * freq;
>>>
>>>          DBG("%s: count=%d, timeout=%d, freq=%lu\n",
>>> @@ -201,20 +201,20 @@ static int s3c2410wdt_set_heartbeat(struct
>>> watchdog_device *wdd, unsigned timeou
>>>
>>>          if (count >= 0x10000) {
>>>                  for (divisor = 1; divisor <= 0x100; divisor++) {
>>> -                       if ((count / divisor) < 0x10000)
>>> +                       if (DIV_ROUND_UP(count, divisor) < 0x10000)
>>>                                  break;
>>>                  }
>>>
>> Since you are at it,
>>          divisor = DIV_ROUND_UP(count + 1, 0x10000);
>> might be faster, simpler, and easier to understand than the loop.
>
> Way to see the forest for the trees!
>
> Your math ends up with a slightly different result than the old code,
> though.  One example is when the count is 0x1ffff.  You'll end up with
> a divider of 2 and I'll end up with a divider of 3.
>
> I think we just want:
>
> divisor = DIV_ROUND_UP(count, 0xffff);
>
> ...that produces the same result as the old loop, but am curious to
> know why you chose the "count + 1" and "0x10000".
>

Hi Doug,

I thought the idea was to keep (count / div) less than 0x10000, which you get
by dividing through 0x10000. 0x10000 / 0x10000 = 1, though, so I added 1
to the counter. But maybe I was thinking too much ;-).

Now, 0x1ffff / 2 = 0xffff is still lower than 0x10000, which is what
I thought is the requirement. Ultimately the error is small either way,
so DIV_ROUND_UP(count, 0xffff) is just as good to me to avoid the loop.

Thanks,
Guenter

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