On 10/19, Mikulas Patocka wrote:
>
> synchronize_rcu() is way slower than msleep(1) -
This depends, I guess. but this doesn't mmatter,
> so I don't see a reason
> why should it be complicated to avoid msleep(1).
I don't think this really needs complications. Please look at this
patch for example. Or initial (single writer) version below. It is
not finished and lacks the barriers too, but I do not think it is
more complex.
Oleg.
struct brw_sem {
long __percpu *read_ctr;
wait_queue_head_t read_waitq;
struct mutex writer_mutex;
struct task_struct *writer;
};
int brw_init(struct brw_sem *brw)
{
brw->writer = NULL;
mutex_init(&brw->writer_mutex);
init_waitqueue_head(&brw->read_waitq);
brw->read_ctr = alloc_percpu(long);
return brw->read_ctr ? 0 : -ENOMEM;
}
void brw_down_read(struct brw_sem *brw)
{
for (;;) {
bool done = false;
preempt_disable();
if (likely(!brw->writer)) {
__this_cpu_inc(*brw->read_ctr);
done = true;
}
preempt_enable();
if (likely(done))
break;
__wait_event(brw->read_waitq, !brw->writer);
}
}
void brw_up_read(struct brw_sem *brw)
{
struct task_struct *writer;
preempt_disable();
__this_cpu_dec(*brw->read_ctr);
writer = ACCESS_ONCE(brw->writer);
if (unlikely(writer))
wake_up_process(writer);
preempt_enable();
}
static inline long brw_read_ctr(struct brw_sem *brw)
{
long sum = 0;
int cpu;
for_each_possible_cpu(cpu)
sum += per_cpu(*brw->read_ctr, cpu);
return sum;
}
void brw_down_write(struct brw_sem *brw)
{
mutex_lock(&brw->writer_mutex);
brw->writer = current;
synchronize_sched();
/*
* Thereafter brw_*_read() must see ->writer != NULL,
* and we should see the result of __this_cpu_inc().
*/
for (;;) {
set_current_state(TASK_UNINTERRUPTIBLE);
if (brw_read_ctr(brw) == 0)
break;
schedule();
}
__set_current_state(TASK_RUNNING);
/*
* We can add another synchronize_sched() to avoid the
* spurious wakeups from brw_up_read() after return.
*/
}
void brw_up_write(struct brw_sem *brw)
{
brw->writer = NULL;
synchronize_sched();
wake_up_all(&brw->read_waitq);
mutex_unlock(&brw->writer_mutex);
}
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