Commit 4192f672 authored by Stefan Hajnoczi's avatar Stefan Hajnoczi Committed by David S. Miller

vsock: make listener child lock ordering explicit

There are several places where the listener and pending or accept queue
child sockets are accessed at the same time.  Lockdep is unhappy that
two locks from the same class are held.

Tell lockdep that it is safe and document the lock ordering.

Originally Claudio Imbrenda <imbrenda@linux.vnet.ibm.com> sent a similar
patch asking whether this is safe.  I have audited the code and also
covered the vsock_pending_work() function.
Suggested-by: default avatarClaudio Imbrenda <imbrenda@linux.vnet.ibm.com>
Signed-off-by: default avatarStefan Hajnoczi <stefanha@redhat.com>
Signed-off-by: default avatarDavid S. Miller <davem@davemloft.net>
parent 48f1dcb5
......@@ -61,6 +61,14 @@
* function will also cleanup rejected sockets, those that reach the connected
* state but leave it before they have been accepted.
*
* - Lock ordering for pending or accept queue sockets is:
*
* lock_sock(listener);
* lock_sock_nested(pending, SINGLE_DEPTH_NESTING);
*
* Using explicit nested locking keeps lockdep happy since normally only one
* lock of a given class may be taken at a time.
*
* - Sockets created by user action will be cleaned up when the user process
* calls close(2), causing our release implementation to be called. Our release
* implementation will perform some cleanup then drop the last reference so our
......@@ -443,7 +451,7 @@ void vsock_pending_work(struct work_struct *work)
cleanup = true;
lock_sock(listener);
lock_sock(sk);
lock_sock_nested(sk, SINGLE_DEPTH_NESTING);
if (vsock_is_pending(sk)) {
vsock_remove_pending(listener, sk);
......@@ -1292,7 +1300,7 @@ static int vsock_accept(struct socket *sock, struct socket *newsock, int flags)
if (connected) {
listener->sk_ack_backlog--;
lock_sock(connected);
lock_sock_nested(connected, SINGLE_DEPTH_NESTING);
vconnected = vsock_sk(connected);
/* If the listener socket has received an error, then we should
......
Markdown is supported
0%
or
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment