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

kernel/irq/manage.c

@@ -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);