Corey Minyard <minyard@xxxxxxx> wrote:With even more context:
diff -puN net/ipmi/af_ipmi.c~af_ipmi-locking-fix net/ipmi/af_ipmi.c>--- 25/net/ipmi/af_ipmi.c~af_ipmi-locking-fix Tue Feb 24 16:56:36 2004
>+++ 25-akpm/net/ipmi/af_ipmi.c Tue Feb 24 16:57:00 2004
>@@ -336,6 +336,7 @@ static int ipmi_recvmsg(struct kiocb *io
> }
> > timeo = ipmi_wait_for_queue(i, timeo);
>+ spin_lock_irqsave(&i->lock, flags);
> }
> > rcvmsg = list_entry(i->msg_list.next, struct ipmi_recv_msg, link);
>
>
> which may or may not be correct.
>
Actually, I believe the code is correct, and your change will break it. This is in a "while (1)" loop, and the only way to get out of this loop is to return with the lock not held or to break out of the loop with the lock held (and later code will unlock it). Am I correct here?
With a little more context:
+ spin_unlock_irqrestore(&i->lock, flags);
+ if (!timeo) {
+ return -EAGAIN;
+ } else if (signal_pending (current)) {
+ dbg("Signal pending: %d", 1);
+ return -EINTR;
+ }
+
+ timeo = ipmi_wait_for_queue(i, timeo);
+ }
+
+ rcvmsg = list_entry(i->msg_list.next, struct ipmi_recv_msg, link);
+ list_del(&rcvmsg->link);
+ spin_unlock_irqrestore(&i->lock, flags);
See, there's a direct code path from one spin_unlock() to the other. And
ipmi_wait_for_queue() does not retake the lock.