Date: Fri, 3 Jun 2005 15:24:19 +0200 From: "Norbert Koch" <NKoch@demig.de> To: "Freebsd-Hackers@Freebsd. Org" <freebsd-hackers@freebsd.org> Subject: synchronization question about /sys/dev/vkbd/vkbd.c Message-ID: <000001c5683f$8c7c1e60$4801a8c0@ws-ew-3.W2KDEMIG>
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Hello.
When looking at /sys/dev/vkbd/vkbd.c I found
one thing, that I do not understand.
There are three places, where a flag TASK is used:
1. in vkbd_dev_close():
while(state->ks_flag & TASK) VKBD_SLEEP (...);
2. in vkbd_dev_write()
VKBD_LOCK ();
...
if (!(state->ks_flags & TASK) && task_enqueue(...))
state->ks_flags |= TASK;
...
VKBD_UNLOCK ();
3. in vkbd_dev_intr()
...
/* call intr */
...
VKBD_LOCK();
state->ks_flags &= ~TASK;
wakeup(...);
VKBD_UNLOCK();
As I understand:
vkbd_dev_write() writes data into its queue
and wants vkbd_dev_intr() to process the queue.
My question is:
Is it not possible, that vkbd_dev_intr() could be
interrupted at any position before the VKBD_LOCK()
and then vkbd_dev_write() called?
If yes, how should vkbd_dev_write() know, that it should
call task_enqueue(), as TASK is still set?
Why not always call task_enqueue() unconditionally here
and leave TASK only to synchronize the close call?
Thank you for any help,
Norbert Koch
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