There's a bug in irq_alloc_virt() if it's asked for more than 1 interrupt,
if it can't find a slot it might look past the end of the irq_map.
To be clear: the bug is that the continue affects the inner for loop,
not the outer one, so i becomes j + 1 and then we continue the inner
loop without checking if i is still <= limit.
This fixes it. No one in the kernel actually calls this with count >
1, so it's not critical.
Signed-off-by: Michael Ellerman <michael@ellerman.id.au>
Signed-off-by: Benjamin Herrenschmidt <benh@kernel.crashing.org>
Signed-off-by: Paul Mackerras <paulus@samba.org>
{
unsigned long flags;
unsigned int i, j, found = NO_IRQ;
- unsigned int limit = irq_virq_count - count;
if (count == 0 || count > (irq_virq_count - NUM_ISA_INTERRUPTS))
return NO_IRQ;
/* Look for count consecutive numbers in the allocatable
* (non-legacy) space
*/
- for (i = NUM_ISA_INTERRUPTS; i <= limit; ) {
- for (j = i; j < (i + count); j++)
- if (irq_map[j].host != NULL) {
- i = j + 1;
- continue;
- }
- found = i;
- break;
+ for (i = NUM_ISA_INTERRUPTS, j = 0; i < irq_virq_count; i++) {
+ if (irq_map[i].host != NULL)
+ j = 0;
+ else
+ j++;
+
+ if (j == count) {
+ found = i - count + 1;
+ break;
+ }
}
if (found == NO_IRQ) {
spin_unlock_irqrestore(&irq_big_lock, flags);