Re: [PATCH v5 1/2] Use "request_muxed_region" in it87 watchdog drivers

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On Wed, 13 Apr 2011 01:05:33 -0700, Natarajan Gurumoorthy wrote:
> Hans,
>       Comments below
> 
> On Tue, Apr 12, 2011 at 11:50 PM, Hans de Goede <hdegoede@xxxxxxxxxx> wrote:
> >
> > You shouldn't (void) this, there is a reason you get a warning
> > otherwise! request_muxed_region can still fail if some other driver
> > has done a none muxed request_region on the same region, and in that
> > case you should not continue with accessing the io-ports.
> There are 3 it87 drivers and they all have to do the exact same
> sequence to put the chip into a mode where they can modify its state.
> The sequences involve non atomic sequences that write locations 0x2e
> and 0x2f. When they are done they write a different sequence to these
> 2 locations. The entry routine is superio_enter and exit is
> superio_exit. All the it87 drivers reserve these 2 locations before
> they start manipulating the chip. This macro will hold off requestors
> if the resource is busy because one of the other drivers is
> manipulating the chip. Once the  an it87 driver is done it calls
> superio_exit which will release the reservation on those 2 locations
> letting any other driver on the wait queue to now gain access two
> locations.
> 
> Please read code in kernel/resource.c function "__request_region".
> "request_muxed_region" turns on IORESOURCE_MUXED bit and that means
> that only  way an it87 driver will get back from a call to
> "request_muxed_region" is when it gets hold of the region.
> 
> The scenario you mention above can never happen.

Let me be straight clear, as apparently you have difficulties
understanding Hans's simple request:

You do not get to (void) the return of request_muxed_region(), period.
This is _not_ negotiable.

What other it87 drivers currently in the tree do or don't do is totally
irrelevant. There can be new it87 drivers added later, there can be
out-of-tree it87 drivers (including old copies of in-tree ones), and
there can be non-it87 drivers accessing the I/O ports (or at least
attempting to.) So the scenario mentioned by Hans can very well happen,
and you have to deal with it.

Thanks,
-- 
Jean Delvare

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