Would this work? Just call getLock() in a while(! getLock()) {} loop
until you get it.
boolean getLock() {
String lockIndex = "LockIndex" +
MemcacheService.increment("descriptiveLockName", 1, 0);
MemcacheService.put("DescriptiveLockName",
lockIndex,
null,
MemcacheService.SetPolicy.ADD_ONLY_IF_NOT_PRESENT);
if(memcacheService.get("DescriptiveLockName") == lockIndex) {
// Yay, got the lock.
return lockIndex;
}
else {
return false;
}
}
void releaseLock() {
MemcacheService.delete("DescriptiveLockName");
}
On Dec 14, 5:07 am, Andrei Cosmin Fifiiţă <[email protected]>
wrote:
> I was thinking about that (increment/decrement)...
> But, first, if the tasks read the value of the variable that needs to be
> incremented, isn't there a posibility that all tasks will read the same
> value at the same time and then increment it in the same time ?
>
> On 14 December 2010 11:59, Simon <[email protected]> wrote:
>
>
>
>
>
>
>
> > The synchronized block won't work at all, since it's not guaranteed
> > that you only have a single instance of your application running at
> > any point in time. As soon as you have multiple instances then you
> > will be synchronizing in different JVMs and hence you'll get multiple
> > threads accessing the cache, and hence the same problem will occur.
>
> > You could probably use the Memcache increment/decrement functionality
> > to hook together a rough cross-instance synchronization, since they
> > increment and decrement atomically.
>
> > On Dec 14, 5:22 am, Michael Weinberg <[email protected]> wrote:
> > > You can use the standard Java synchronization to synchronize the task
> > > threads, e.g.
>
> > > synchronized (YourMemcachedDataClass.class) {
> > > YourMemcachedDataClass cachedData =
> > > (YourMemcachedDataClass)cache.get(CACHE_KEY);
>
> > > if (cachedData == null)
> > > cachedData = new YourMemcachedDataClass ();
>
> > > cachedData.add(...);
> > > cachedData.set(...);
>
> > > cache.put(CACHE_KEY, cachedData);
>
> > > }
>
> > > this way only 1 thread at a time will get into the synchronized block.
>
> > > Hope it helps..
>
> > > Michael Weinberg
>
> > > On Dec 13, 11:41 am, Ice13ill <[email protected]> wrote:
>
> > > > Maybe there's something that i don't know very well about the memcache
> > > > service and this problem is simpler then i thing it is... If so,
> > > > please advice :)
>
> > > > On Dec 13, 6:38 pm, Ice13ill <[email protected]> wrote:
>
> > > > > I need advice for implementing concurrent access to a memcache
> > > > > variable, when used by multiple running tasks.
> > > > > The problem is this: I have a number of tasks reading from datastore
> > > > > and doing some processing.
> > > > > When the processing is complete, i add a String to a List stored in
> > > > > memcache. So i need to get the List from memcache, add a new element,
> > > > > and put it back. The pb is that if 2 tasks write on that variable, i
> > > > > get an exception (I understand that the memcache service gives me a
> > > > > "clone" of the object, not the object itself, or am I wrong ?)
> > > > > So how can i avoid more than one tasks writing in the same object ?
>
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