genirq: do not leave interupts enabled on free_irq

The default_disable() function was changed in commit:

 76d21601
 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: NMichael Hennerich <Michael.Hennerich@analog.com>
Signed-off-by: NThomas Gleixner <tglx@linutronix.de>
Acked-by: NIngo Molnar <mingo@elte.hu>
Cc: stable@kernel.org
Tested-by: NMichael Hennerich <Michael.Hennerich@analog.com>

kernel/irq/chip.c

@@ -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)