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Message-ID: <20081104222927.GE4641@ovro.caltech.edu>
Date:	Tue, 4 Nov 2008 14:29:27 -0800
From:	Ira Snyder <iws@...o.caltech.edu>
To:	Arnd Bergmann <arnd@...db.de>
Cc:	linuxppc-dev@...abs.org, Stephen Hemminger <shemminger@...tta.com>,
	netdev@...r.kernel.org, linux-kernel@...r.kernel.org,
	Jan-Bernd Themann <THEMANN@...ibm.com>
Subject: Re: [PATCH RFC v2] net: add PCINet driver

On Tue, Nov 04, 2008 at 09:23:03PM +0100, Arnd Bergmann wrote:

Big snip.

> > 
> > I tried to make the locking do only what was needed. I just couldn't get
> > it correct unless I used spin_lock_irqsave(). I was able to get the
> > system to deadlock otherwise. This is why I posted the driver for
> > review, I could use some help here.
> > 
> > It isn't critical anyway. You can always use spin_lock_irqsave(), it is
> > just a little slower, but it will always work :)
> 
> I like the documenting character of the spinlock functions. E.g. if you
> use spin_lock_irq() in a function, it is obvious that interrupts are enabled,
> and if you use spin_lock() on a lock that requires disabling interrupts,
> you know that interrupts are already off.
>  

Ok, I've started addressing your comments. I'm having trouble with the
locking again. Let me see if my understanding is correct:

spin_lock_irqsave() disables interrupts and saves the interrupt state
spin_unlock_irqrestore() MAY re-enable interrupts based on the saved
flags

spin_lock_irq() disables interrupts, and always turns them back on when
spin_unlock_irq() is called

spin_lock_bh() disables softirqs, spin_unlock_bh() re-enables them

spin_lock() and spin_unlock() are just regular spinlocks


So, since interrupts are disabled while my interrupt handler is running,
I think I should be able to use spin_lock() and spin_unlock(), correct?

But sparse gives me the following warning:
wqt.c:185:9: warning: context imbalance in 'wqt_interrupt': wrong count at exit
wqt.c:185:9:    context 'lock': wanted 0, got 1

If I'm using spin_lock_irqsave() and spin_lock_irqrestore() I do not get
the same warnings. Therefore I must have some misunderstanding :)

Thanks,
Ira
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