>From memory.
The code isn't trying to eliminate that race, no. If/when it arises, and all
threads are idle, then the timeout will ensure progress.
In particular, if there is no work for -any- threads, it is desirable for at
least
most to go idle. As work ramps up, threads will be woken.
The next optimization, I would intuit, might be to have a leader that is
resistant
to idle.
Matt
----- "Malahal Naineni" <mala...@us.ibm.com> wrote:
> The nfs_rpc_enqueue_req is called by the producer and
> nfs_rpc_dequeue_req is called by the consumer. Here is the high level
> view of those functions.
>
> In particular, if a consumer finds no request in the queue, and then
> attempts to go to sleep, but before he adds himself to the wait list
> (before step 2b), if producer adds another request at that instant,
> the
> producer would find the wait list empty and couldn't wake up any
> consumer.
>
> Later the consumer would add himself to the wait list and goes to
> sleep
> (in a loop actually due to timedwait call. This will continue if
> there
> was no other request coming in. Am I missing something?
>
> nfs_rpc_dequeue_req()
> {
> 1. request = nfs_rpc_consume_req()
> 2. if (request == NULL) { /* no work to do */
> a) set Wqe_LFlag_WaitSync flag
> b) add itself to the wait list
> c) while not Wqe_LFlag_SyncDone {
> pthread_cond_timedwait(5 seconds)
> }
> goto step 1; /* woke up with SyncDone */
> }
> }
>
> nfs_rpc_enqueue_req()
> {
> 1. add request to the list
> 2. if (wait list is not empty) {
> a) pick up first waiting thread.
> b) set Wqe_LFlag_SyncDone
> c) if Wqe_LFlag_WaitSync() pthread_cond_signal().
> }
> }
>
>
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--
Matt Benjamin
CohortFS, LLC.
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