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genirq: do not leave interupts enabled on free_irq

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The default_disable() function was changed in commit:

 76d2160147
 genirq: do not mask interrupts by default

It removed the mask function in favour of the default delayed
interrupt disabling. Unfortunately this also broke the shutdown in
free_irq() when the last handler is removed from the interrupt for
those architectures which rely on the default implementations. Now we
can end up with a enabled interrupt line after the last handler was
removed, which can result in spurious interrupts.

Fix this by adding a default_shutdown function, which is only
installed, when the irqchip implementation does provide neither a
shutdown nor a disable function.

[@stable: affected versions: .21 - .24 ]

Pointed-out-by: Michael Hennerich <Michael.Hennerich@analog.com>
Signed-off-by: Thomas Gleixner <tglx@linutronix.de>
Acked-by: Ingo Molnar <mingo@elte.hu>
Cc: stable@kernel.org
Tested-by: Michael Hennerich <Michael.Hennerich@analog.com>
This commit is contained in:
Thomas Gleixner 2008-02-18 18:25:17 +01:00
parent 188fd89d53
commit 89d694b9db
1 changed files with 19 additions and 1 deletions
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20
kernel/irq/chip.c
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@ -245,6 +245,17 @@ static unsigned int default_startup(unsigned int irq)
return 0;
}
/*
* default shutdown function
*/
static void default_shutdown(unsigned int irq)
{
struct irq_desc *desc = irq_desc + irq;
desc->chip->mask(irq);
desc->status |= IRQ_MASKED;
}
/*
* Fixup enable/disable function pointers
*/
@ -256,8 +267,15 @@ void irq_chip_set_defaults(struct irq_chip *chip)
chip->disable = default_disable;
if (!chip->startup)
chip->startup = default_startup;
/*
* We use chip->disable, when the user provided its own. When
* we have default_disable set for chip->disable, then we need
* to use default_shutdown, otherwise the irq line is not
* disabled on free_irq():
*/
if (!chip->shutdown)
chip->shutdown = chip->disable;
chip->shutdown = chip->disable != default_disable ?
chip->disable : default_shutdown;
if (!chip->name)
chip->name = chip->typename;
if (!chip->end)