On 09/01, Peter Zijlstra wrote:
>
> > ret = 0;
> >
> > for (;;) {
> > prepare_to_wait_exclusive(wq, &q->wait, mode);
> >
> > if (test_bit(&q->key.bit_nr, &q->key.flag))
> > ret = action(&q->key, mode);
> >
> > if (!test_and_set_bit(&q->key.bit_nr, &q->key.flag)) {
> > /* we got the lock anyway, ignore the signal */
> > ret = 0;
> > break;
> > }
> >
> > if (ret)
> > break;
> > }
> > finish_wait(wq, &q->wait);
> >
> > return ret;
> >
> >
> > Would not that work too?
>
> Nope, because we need to do that finish_wait() before
> test_and_set_bit()..
Yes, I meant
int __sched
__wait_on_bit_lock(wait_queue_head_t *wq, struct wait_bit_queue *q,
wait_bit_action_f *action, unsigned mode)
{
int ret = 0;
for (;;) {
prepare_to_wait_exclusive(wq, &q->wait, mode);
if (test_bit(q->key.bit_nr, q->key.flags))
ret = action(&q->key, mode);
finish_wait(wq, &q->wait);
if (!test_and_set_bit(q->key.bit_nr, q->key.flags))
return 0;
else if (ret)
return ret;
}
}
> Also the problem with doing finish_wait() unconditionally would be
> destroying the FIFO order. With a bit of bad luck you'd get starvation
> cases :/
OK, I didn't think about that, thanks.
Oleg.
