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genirq: Avoid deadlock in spurious handling 

commit 52553ddf(genirq: fix regression in irqfixup, irqpoll)
introduced a potential deadlock by calling the action handler with the
irq descriptor lock held.

Remove the call and let the handling code run even for an interrupt
where only a single action is registered. That matches the goal of
the above commit and avoids the deadlock.

Document the confusing action = desc->action reload in the handling
loop while at it.

Reported-and-tested-by: "Wang, Warner" <warner.wang@hp.com>
Tested-by: Edward Donovan <edward.donovan@numble.net>
Cc: "Wang, Song-Bo (Stoney)" <song-bo.wang@hp.com>
Cc: stable@vger.kernel.org
Signed-off-by: Thomas Gleixner <tglx@linutronix.de>
This commit is contained in:
Thomas Gleixner 2012-11-23 10:08:44 +01:00
parent 04aa530ec0
commit e716efde75
1 changed files with 3 additions and 4 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

7
kernel/irq/spurious.c
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@ -80,13 +80,11 @@ static int try_one_irq(int irq, struct irq_desc *desc, bool force)
/*
* All handlers must agree on IRQF_SHARED, so we test just the
* first. Check for action->next as well.
* first.
*/
action = desc->action;
if (!action || !(action->flags & IRQF_SHARED) ||
(action->flags & __IRQF_TIMER) ||
(action->handler(irq, action->dev_id) == IRQ_HANDLED) ||
!action->next)
(action->flags & __IRQF_TIMER))
goto out;
/* Already running on another processor */
@ -104,6 +102,7 @@ static int try_one_irq(int irq, struct irq_desc *desc, bool force)
do {
if (handle_irq_event(desc) == IRQ_HANDLED)
ret = IRQ_HANDLED;
/* Make sure that there is still a valid action */
action = desc->action;
} while ((desc->istate & IRQS_PENDING) && action);
desc->istate &= ~IRQS_POLL_INPROGRESS;