Don't call "note_interrupt()" with irq descriptor lock held

This reverts commit f72fa70, and solves
the problem that it tried to fix by simply making "__do_IRQ()" call the
note_interrupt() function without the lock held, the way everybody else
does.

It should be noted that all interrupt handling code must never allow the
descriptor actors to be entered "recursively" (that's why we do all the
magic IRQ_PENDING stuff in the first place), so there actually is
exclusion at that much higher level, even in the absense of locking.

Acked-by: Vivek Goyal <vgoyal@in.ibm.com>
Acked-by:Pavel Emelianov <xemul@openvz.org>
Cc: Andrew Morton <akpm@osdl.org>
Cc: Ingo Molnar <mingo@redhat.com>
Cc: Adrian Bunk <bunk@stusta.de>
Signed-off-by: Linus Torvalds <torvalds@osdl.org>

kernel/irq/handle.c

@@ -231,10 +231,10 @@ fastcall unsigned int __do_IRQ(unsigned int irq)
 		spin_unlock(&desc->lock);
 
 		action_ret = handle_IRQ_event(irq, action);
-
-		spin_lock(&desc->lock);
 		if (!noirqdebug)
 			note_interrupt(irq, desc, action_ret);
+
+		spin_lock(&desc->lock);
 		if (likely(!(desc->status & IRQ_PENDING)))
 			break;
 		desc->status &= ~IRQ_PENDING;

kernel/irq/spurious.c

@@ -147,11 +147,7 @@ void note_interrupt(unsigned int irq, struct irq_desc *desc,
 	if (unlikely(irqfixup)) {
 		/* Don't punish working computers */
 		if ((irqfixup == 2 && irq == 0) || action_ret == IRQ_NONE) {
-			int ok;
-
-			spin_unlock(&desc->lock);
-			ok = misrouted_irq(irq);
-			spin_lock(&desc->lock);
+			int ok = misrouted_irq(irq);
 			if (action_ret == IRQ_NONE)
 				desc->irqs_unhandled -= ok;
 		}