Hi...
On Mon, Mar 30, 2009 at 4:55 AM, David Neiss <davidaneiss@xxxxxxxxx> wrote:
> So, having kernel code which does:
>
> set_current_state(TASK_INTERRUPTIBLE);
> schedule();
>
> causes blocking, as expected, and /proc/stat shows increasing idle
> task time and non increasing io wait time. Fine.
>
> Changing schedule() to io_schedule(), time now seems to be split
> between idle task and io_wait time.
>
> Question, why does the system impute any running time to io_wait time?
> The task that got blocked is presumably not doing anything and just
> sitting there inactive until it gets woken up, so I would expect just
> idle time to be increasing. I know its trying to account for io_wait
> time, but how does it know how many jiffies to be adding to io_wait
> time? Googling on io_wait and/or io_schedule are not turning up much.
> So, I'm obviously not understanding things here. Any input?
Interesting, never thought about it before. As usual, lxr help us to
locate the related function:
void __sched io_schedule(void)
{
struct rq *rq = &__raw_get_cpu_var(runqueues);
delayacct_blkio_start();
atomic_inc(&rq->nr_iowait);
schedule();
atomic_dec(&rq->nr_iowait);
delayacct_blkio_end();
}
further checking of delayacct_blkio_start roughly tells me that it
starts to account for io_wait. So that's the answer, this is the
function that does the accounting.
However, due to the way time accounting done in Linux (in every
ticks), there could be imprecisionness.
regards,
Mulyadi.
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