Multiple locks and missing wakeup.

[Adrian Chadd]

If you don't need the lock over the do_stuff, then:

for(;;) {
    list_t lcl;

    init(lcl);

    lock;
    move list1 to lcl;
    unlock;

    while (! list_empty(lcl))
       do_crap_on_lcl_head();
}


-a


On 7 April 2014 23:34, Edward Tomasz Napierała <trasz at freebsd.org> wrote:
> Let's say I have a kernel thread processing elements from a queue,
> sleeping until there is work to do; something like this:
>
> mtx_lock(&mtx1);
> for (;;) {
>         while (!LIST_EMPTY(&list1)) {
>                 elt = LIST_FIRST(&list1);
>                 do_stuff(elt);
>                 LIST_REMOVE(&list1, elt);
>         }
>         sleep(&list1, &mtx1);
> }
> mtx_unlock(&mtx1);
>
> Now, is there some way to make it work with two lists, protected
> by different mutexes?  The mutex part is crucial here; the whole
> point of this is to reduce lock contention on one of the lists.  The
> following code would result in a missing wakeup:
>
> mtx_lock(&mtx1);
> for (;;) {
>         while (!LIST_EMPTY(&list1)) {
>                 elt = LIST_FIRST(&list1);
>                 do_stuff(elt);
>                 LIST_REMOVE(&list1, elt);
>         }
>
>         mtx_lock(&mtx2);
>         while (!LIST_EMPTY(&list2)) {
>                 elt = LIST_FIRST(&list2);
>                 do_other_stuff(elt);
>                 LIST_REMOVE(&list2, elt);
>         }
>         mtx_unlock(&mtx2);
>
>         sleep(&list1, &mtx1);
> }
> mtx_unlock(&mtx1);
>
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