linux-kernel - Re: [PATCH v2 1/3] watchdog: qcom: add support for KPSS WDT

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Message-ID: <20140924183050.GI868@joshc.qualcomm.com>
Date:	Wed, 24 Sep 2014 13:30:50 -0500
From:	Josh Cartwright <joshc@...eaurora.org>
To:	Guenter Roeck <linux@...ck-us.net>
Cc:	Wim Van Sebroeck <wim@...ana.be>, linux-watchdog@...r.kernel.org,
	Grant Likely <grant.likely@...aro.org>,
	Rob Herring <robh+dt@...nel.org>,
	linux-arm-msm@...r.kernel.org,
	linux-arm-kernel@...ts.infradead.org,
	Kumar Gala <galak@...eaurora.org>,
	linux-kernel@...r.kernel.org, devicetree@...r.kernel.org
Subject: Re: [PATCH v2 1/3] watchdog: qcom: add support for KPSS WDT

On Wed, Sep 24, 2014 at 08:58:54AM -0700, Guenter Roeck wrote:
> On Tue, Sep 23, 2014 at 06:04:36PM -0500, Josh Cartwright wrote:
> > Add a driver for the watchdog timer block found in the Krait Processor
> > Subsystem (KPSS) on the MSM8960, APQ8064, and IPQ8064.
> > 
> > Signed-off-by: Josh Cartwright <joshc@...eaurora.org>
> 
> Hi Josh,
> 
> looks much better. Couple of comments inline.

Thanks for another review!

[..]
> > +++ b/drivers/watchdog/Kconfig
> > @@ -443,6 +443,19 @@ config TEGRA_WATCHDOG
> >  	  To compile this driver as a module, choose M here: the
> >  	  module will be called tegra_wdt.
> >  
> > +config QCOM_WDT
> > +	tristate "QCOM watchdog"
> > +	depends on HAS_IOMEM
> > +	depends on ARCH_QCOM
> > +	select WATCHDOG_CORE
> > +	help
> > +	  Say Y here to include Watchdog timer support for the watchdog found
> > +	  on QCOM chipsets.  Currently supported targets are the MSM8960,
> > +	  APQ8064, and IPQ8064.
> > +
> > +	  To compile this driver as a module, choose M here: the
> > +	  module will be called qcom_wdt.
> > +
> >  # AVR32 Architecture
[..]
> > +static int qcom_wdt_probe(struct platform_device *pdev)
> > +{
> > +	struct qcom_wdt *wdt;
> > +	struct resource *res;
> > +	unsigned long freq;
> > +	int ret;
> > +
> > +	wdt = devm_kzalloc(&pdev->dev, sizeof(*wdt), GFP_KERNEL);
> > +	if (!wdt)
> > +		return -ENOMEM;
> > +
> > +	res = platform_get_resource(pdev, IORESOURCE_MEM, 0);
> > +	wdt->base = devm_ioremap_resource(&pdev->dev, res);
> > +	if (IS_ERR(wdt->base))
> > +		return PTR_ERR(wdt->base);
> > +
> > +	wdt->clk = devm_clk_get(&pdev->dev, NULL);
> > +	if (IS_ERR(wdt->clk))
> > +		return PTR_ERR(wdt->clk);
> > +
> > +	ret = clk_prepare_enable(wdt->clk);
> > +	if (ret) {
> > +		dev_err(&pdev->dev, "failed to setup clock\n");
> > +		return ret;
> > +	}
> > +
> > +	/*
> > +	 * We use the clock rate to calculate the max timeout, so ensure it's
> > +	 * not zero to avoid a divide-by-zero exception.
> > +	 */
> > +	freq = clk_get_rate(wdt->clk);
> > +	if (freq == 0) {
> > +		dev_err(&pdev->dev, "invalid clock rate\n");
> > +		return -EINVAL;
> > +	}
> 
> This will need clk_disable_unprepare().

Yep.  Nice catch.  Will fix.

> Since you are reading the frequency here, it might make sense to store it
> in struct qcom_wdt so you don't have to call clk_get_rate() again in the
> start function.

Yeah, it doesn't save much, but I'll go ahead and add it.

> > +	wdt->wdd.dev = &pdev->dev;
> > +	wdt->wdd.info = &qcom_wdt_info;
> > +	wdt->wdd.ops = &qcom_wdt_ops;
> > +	wdt->wdd.min_timeout = 1;
> > +	wdt->wdd.max_timeout = 0x10000000U / freq;
>
> What if the frequency turns out to be larger than 8947848 Hz ?
> Then your maximum timeout is below the default timeout.
> And if the frequency is larger than 268435456 Hz, the maximum
> timeout would be 0.

Yes, I should be doing more sanity checking.  I'll do so.

> > +	/*
> > +	 * If 'timeout-sec' unspecified in devicetree, assume a 30 second
> > +	 * default.
> > +	 */
> > +	if (watchdog_init_timeout(&wdt->wdd, 0, &pdev->dev))
> > +		wdt->wdd.timeout = 30;
> 
> You can just initialize timeout above with 30 seconds. Saves you the if
> statement here.

Great.  Thanks.

> > +	ret = watchdog_register_device(&wdt->wdd);
> > +	if (ret) {
> > +		dev_err(&pdev->dev, "failed to register watchdog\n");
>
> This will need a clk_disable_unprepare().
>
> Given that this is needed twice, you might want to consider using
> error exit code below, as suggested in CodingStyle.

Indeed.  Will do.

Thanks again,
  Josh

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