m68k/mac: Fix apparent race condition in Baboon interrupt dispatch

The algorithm used in baboon_irq() appears to be subject to a race
condition: an IRQ flag could be lost if asserted between the MOV
instructions from and to the interrupt flag register. However,
testing shows that the write to the flag register has no effect.
Rewrite this loop to remove the apparent race condition.

No-one seems to know how to clear Baboon IRQ flags, or whether
that's even possible, so add a comment about this.

Tested-by: Stan Johnson <userm57@yahoo.com>
Signed-off-by: Finn Thain <fthain@telegraphics.com.au>
Signed-off-by: Geert Uytterhoeven <geert@linux-m68k.org>
This commit is contained in:
Finn Thain 2018-01-27 18:51:40 -05:00 committed by Geert Uytterhoeven
parent e5f0d2e2a1
commit a1eb1cdf4e

View File

@ -18,10 +18,6 @@
int baboon_present;
static volatile struct baboon *baboon;
#if 0
extern int macide_ack_intr(struct ata_channel *);
#endif
/*
* Baboon initialization.
*/
@ -41,33 +37,26 @@ void __init baboon_init(void)
}
/*
* Baboon interrupt handler. This works a lot like a VIA.
* Baboon interrupt handler.
* XXX how do you clear a pending IRQ? is it even necessary?
*/
static void baboon_irq(struct irq_desc *desc)
{
int irq_bit, irq_num;
unsigned char events;
short events, irq_bit;
int irq_num;
events = baboon->mb_ifr & 0x07;
if (!events)
return;
irq_num = IRQ_BABOON_0;
irq_bit = 1;
do {
if (events & irq_bit) {
baboon->mb_ifr &= ~irq_bit;
if (events & irq_bit) {
events &= ~irq_bit;
generic_handle_irq(irq_num);
}
++irq_num;
irq_bit <<= 1;
irq_num++;
} while(events >= irq_bit);
#if 0
if (baboon->mb_ifr & 0x02) macide_ack_intr(NULL);
/* for now we need to smash all interrupts */
baboon->mb_ifr &= ~events;
#endif
} while (events);
}
/*