提交 b42172fc 编写于 作者: L Linus Torvalds

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

This reverts commit f72fa707, 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: NVivek Goyal <vgoyal@in.ibm.com>
Acked-by: NPavel 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: NLinus Torvalds <torvalds@osdl.org>
上级 0f9005a6
......@@ -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;
......
......@@ -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;
}
......
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