"adrian(a)jboss.org" wrote : | Yes. But you're welcome to examine the code and try to come up with new testcases | where it breaks. The only thing that should be not supported is loops. :-) | See DeployersImpl.sort() Do we really need to check all current deployers (O(n^2))? Since the original set is already ordered, thus we only need to find the position of the newly added deployer (O(n)). View the original post : http://www.jboss.com/index.html?module=bb&op=viewtopic&p=4059954#... Reply to the post : http://www.jboss.com/index.html?module=bb&op=posting&mode=reply&a...