leandrof/linux
1
0
Fork 0
forked from Minki/linux
Code Issues Pull Requests Packages Projects Releases Wiki Activity

fanotify: dont merge permission events

Browse Source 
Boyd Yang reported a problem for the case that multiple threads of the same
thread group are waiting for a reponse for a permission event.
In this case it is possible that some of the threads are never woken up, even
if the response for the event has been received
(see http://marc.info/?l=linux-kernel&m=131822913806350&w=2).

The reason is that we are currently merging permission events if they belong to
the same thread group. But we are not prepared to wake up more than one waiter
for each event. We do

wait_event(group->fanotify_data.access_waitq, event->response ||
			atomic_read(&group->fanotify_data.bypass_perm));
and after that
  event->response = 0;

which is the reason that even if we woke up all waiters for the same event
some of them may see event->response being already set 0 again, then go back to
sleep and block forever.

With this patch we avoid that more than one thread is waiting for a response
by not merging permission events for the same thread group any more.

Reported-by: Boyd Yang <boyd.yang@gmail.com>
Signed-off-by: Lino Sanfilippo <LinoSanfilipp@gmx.de>
Signed-off-by: Eric Paris <eparis@redhat.com>
This commit is contained in:
Lino Sanfilippo 2012-03-23 02:42:23 +01:00 committed by Eric Paris
parent 0a6b6bd591
commit 03a1cec1f1
1 changed files with 6 additions and 0 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

6
fs/notify/fanotify/fanotify.c
View File

@ -18,6 +18,12 @@ static bool should_merge(struct fsnotify_event *old, struct fsnotify_event *new)
old->tgid == new->tgid) {
switch (old->data_type) {
case (FSNOTIFY_EVENT_PATH):
#ifdef CONFIG_FANOTIFY_ACCESS_PERMISSIONS
/* dont merge two permission events */
if ((old->mask & FAN_ALL_PERM_EVENTS) &&
(new->mask & FAN_ALL_PERM_EVENTS))
return false;
#endif
if ((old->path.mnt == new->path.mnt) &&
(old->path.dentry == new->path.dentry))
return true;