linux-kernel - Re: [PATCH 1/5] soc: mediatek: Add infracfg misc driver support

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Message-ID: <20150519074535.GY6325@pengutronix.de>
Date:	Tue, 19 May 2015 09:45:35 +0200
From:	Sascha Hauer <s.hauer@...gutronix.de>
To:	Daniel Kurtz <djkurtz@...omium.org>
Cc:	"linux-arm-kernel@...ts.infradead.org" 
	<linux-arm-kernel@...ts.infradead.org>,
	"open list:OPEN FIRMWARE AND..." <devicetree@...r.kernel.org>,
	Kevin Hilman <khilman@...nel.org>,
	"linux-kernel@...r.kernel.org" <linux-kernel@...r.kernel.org>,
	linux-mediatek@...ts.infradead.org,
	Sasha Hauer <kernel@...gutronix.de>,
	Matthias Brugger <matthias.bgg@...il.com>
Subject: Re: [PATCH 1/5] soc: mediatek: Add infracfg misc driver support

On Tue, May 19, 2015 at 02:54:41PM +0800, Daniel Kurtz wrote:
> >> > +       while (1) {
> >> > +               ret = regmap_read(infracfg, INFRA_TOPAXI_PROTECTSTA1, &val);
> >> > +               if (ret)
> >> > +                       return ret;
> >> > +
> >> > +               if ((val & mask) == mask)
> >> > +                       break;
> >> > +
> >> > +               cpu_relax();
> >> > +               if (time_after(jiffies, expired))
> >> > +                       return -EIO;
> >>
> >> I think we should check for timeout first, and then cpu_relax() if
> >> there is still time left (here and in
> >> mtk_infracfg_clear_bus_protection()).  Otherwise we end up doing one
> >> final cpu_relax() without rechecking the register we are polling
> >> (again, I have the same comment for the timeout loops in mtk-scpsys).
> >
> > I think cpu_relax() delays execution in the order of microseconds (I
> > don't actually know, just a guess), so if the timeout is a second the
> > order doesn't really matter. What can happen though is an interrupt
> > after the (val & mask) test but before the timeout check. So to be
> > truly correct we have to repeat the (val & mask) test after the
> > time_after() check. Is that what you want?
> 
> I'm not following, why would you need to repeat (val & mask) test
> after time_after?
> What does an interrupt have to do with it?
> Can you show a code snippet with what exactly you are proposing?

Consider you have this timeout loop:

	while (1) {
		if (success())
			break;

		if (time_after(jiffies, expired))
			return -ETIMEDOUT;
	}

Now when an interupt comes in between success() and time_after() then it
can happen that the delay caused by the interrupt makes the code timeout
even though success() might have become true in the meantime. So to be
correct you have to:

	while (1) {
		if (success())
			break;

		if (time_after(jiffies, expired)) {
			if (success())
				break;
			return -ETIMEDOUT;
	}

Or, if you don't want to repeat the termination condition:

	bool timeout = false;

	while (1) {
		if (success())
			break;

		if (timeout)
			return -ETIMEDOUT;

		if (time_after(jiffies, expired))
			timeout = true;
	}

Anyway, with the timeout of one second used here this is all academic.

Sascha

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