After the removal of epmutex, it's possible eventpoll_release_file()
and ep_free() will try to remove the same epi. The removal of epi
is protected by ep->mtx, so there are two possible sequences:
(1) eventpoll_release_file() first, then ep_free()
there is no double-free, because after the invocation of
eventpoll_release_file(), the epi will be removed from ep->rbr
and ep_free() will not be able to free it.
(2) ep_free() first, then eventpoll_release_file()
double-free is possible because before the invocation of ep_remove()
in ep_free() eventpoll_release_file may has already got the epi from
file->f_ep_links. To protect against the double-free case, check
rb_first_cached() in eventpoll_release_file() to ensure the epi
has not been removed by ep_free()
Signed-off-by: Hou Tao <hout...@huawei.com>
---
fs/eventpoll.c | 13 +++++++++++--
1 file changed, 11 insertions(+), 2 deletions(-)
diff --git a/fs/eventpoll.c b/fs/eventpoll.c
index 27b743b..cd7a9f4 100644
--- a/fs/eventpoll.c
+++ b/fs/eventpoll.c
@@ -1058,13 +1058,22 @@ void eventpoll_release_file(struct file *file)
break;
ep = eventpoll_get_ep(epi->ep);
- /* Current epi has been removed by ep_free() */
+ /* Current epi had been removed by ep_free() */
if (!ep)
continue;
rcu_read_unlock();
mutex_lock_nested(&ep->mtx, 0);
- ep_remove(ep, epi);
+ /*
+ * If rb_first_cached() returns NULL, it means that the current
epi
+ * had been removed by ep_free(). To prevent epi from
double-freeing,
+ * check the condition before invoking ep_remove().
+ * If eventpoll_release_file() frees epi firstly, the epi will
not
+ * be freed again because the epi must have been removed from
ep->rbr
+ * when ep_free() is invoked.
+ */
+ if (rb_first_cached(&ep->rbr))
+ ep_remove(ep, epi);
mutex_unlock(&ep->mtx);
eventpoll_put_ep(ep);
--
2.7.5
