Date: Tue, 8 Apr 2014 11:51:56 +0300 From: Konstantin Belousov <kostikbel@gmail.com> To: Edward Tomasz Napiera?a <trasz@FreeBSD.org> Cc: hackers@freebsd.org Subject: Re: Multiple locks and missing wakeup. Message-ID: <20140408085156.GR21331@kib.kiev.ua> In-Reply-To: <0D69A6A8-43D1-41FB-8C2D-00F5CAD9C86E@FreeBSD.org> References: <0D69A6A8-43D1-41FB-8C2D-00F5CAD9C86E@FreeBSD.org>
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On Tue, Apr 08, 2014 at 08:34:30AM +0200, Edward Tomasz Napiera?a 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_unlock(&mtx1);, right ?
>
> 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);
You should clarify your intent. Do you need to sleep waiting for the work
items to appear for list2 ? What wakeup do you miss in the pseudo-code
you posted ?
If you do need the notification for the list2, use &list1 as the wakeup
channel for it.
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