genirq: Prevent chip buslock deadlock

If a interrupt chip utilizes chip->buslock then free_irq() can
deadlock in the following way:

CPU0				CPU1
				interrupt(X) (Shared or spurious)
free_irq(X)			interrupt_thread(X)
chip_bus_lock(X)
				   irq_finalize_oneshot(X)
				     chip_bus_lock(X)
synchronize_irq(X)
	
synchronize_irq() waits for the interrupt thread to complete,
i.e. forever.

Solution is simple: Drop chip_bus_lock() before calling
synchronize_irq() as we do with the irq_desc lock. There is nothing to
be protected after the point where irq_desc lock has been released.

This adds chip_bus_lock/unlock() to the remove_irq() code path, but
that's actually correct in the case where remove_irq() is called on
such an interrupt. The current users of remove_irq() are not affected
as none of those interrupts is on a chip which requires buslock.

Reported-by: Fredrik Markström <fredrik.markstrom@gmail.com>
Signed-off-by: Thomas Gleixner <tglx@linutronix.de>
Cc: stable@vger.kernel.org
This commit is contained in:
Thomas Gleixner 2015-12-13 18:12:30 +01:00
parent 9f9499ae8e
commit abc7e40c81
1 changed files with 3 additions and 3 deletions

View File

@ -1434,6 +1434,7 @@ static struct irqaction *__free_irq(unsigned int irq, void *dev_id)
if (!desc)
return NULL;
chip_bus_lock(desc);
raw_spin_lock_irqsave(&desc->lock, flags);
/*
@ -1447,7 +1448,7 @@ static struct irqaction *__free_irq(unsigned int irq, void *dev_id)
if (!action) {
WARN(1, "Trying to free already-free IRQ %d\n", irq);
raw_spin_unlock_irqrestore(&desc->lock, flags);
chip_bus_sync_unlock(desc);
return NULL;
}
@ -1475,6 +1476,7 @@ static struct irqaction *__free_irq(unsigned int irq, void *dev_id)
#endif
raw_spin_unlock_irqrestore(&desc->lock, flags);
chip_bus_sync_unlock(desc);
unregister_handler_proc(irq, action);
@ -1553,9 +1555,7 @@ void free_irq(unsigned int irq, void *dev_id)
desc->affinity_notify = NULL;
#endif
chip_bus_lock(desc);
kfree(__free_irq(irq, dev_id));
chip_bus_sync_unlock(desc);
}
EXPORT_SYMBOL(free_irq);