blob: 935dadbda9c3928be3759d9a2c217642fa40c755 (
plain) (
tree)
|
|
Some notes about locking in dnbd3
The order of aquiring multiple locks is
VERY IMPORTANT, as you'll produce a possible deadlock
if you do it in the wrong order.
Take very good care of locking order if you have lots
of functions that call each other. You might lose
track of what's going on. ;)
===== FUSE =====
mutexInit
newAltLock
altLock
connection.sendMutex
requests.lock
===== SERVER =====
This is a list of used locks, in the order they
have to be aquired if you must hold multiple locks:
remoteCloneLock | reloadLock
_clients_lock
_clients[].lock
integrityQueueLock
_images_lock
_images[].lock
pendingLockConsume
pendingLockProduce
uplink.queueLock
altServersLock
client.sendMutex
client.statsLock
statisticsSentLock
statisticsReceivedLock
uplink.rttLock
If you need to lock multiple clients/images/... at once,
lock the client with the lowest array index first.
If the program logic would require to aquire the
locks in a different order, you HAVE TO rework the
code.
For example, if you hold the lock for client 10 and
you need to look up some other client. You MUST NOT
simply fetch the _clients_lock now and then iterate
over the clients until you find the one you need,
as it violates the above order to first lock on the
clients array and then the clients lock.
Instead, you need to release client 10's lock,
then lock on _clients_lock and iterate over the
clients. Now you check if you either encounter
the client you originally held the lock on, or
the client you are looking for. You immediately
lock on those two. You can then release the
_clients_lock and work with both clients.
pseudo code:
// client10 is assumed to be a pointer to
// a client, which happens to be at index 10
lock (client10->lock);
....
// oh, i need another client
unlock(client10->lock);
lock(_clients_lock);
client clientA = NULL, clientB = NULL;
for (i = 0; i < _num_clients; ++i) {
if (client[i] == client10) {
clientA = client[i];
lock(clientA.lock);
} else if (client[i].something == <whatever>) {
clientB = client[i];
lock(clientB.lock);
}
}
unlock(_clients_lock);
if (clientA && clientB) { // Make sure we actually found both!
// DO something important with both clients
}
if (clientA) unlock(clientA.lock);
if (clientB) unlock(clientB.lock);
|
