def join(self, newnode) :
# Base case: First node
if self.__start is None :
self.__start = newnode
self.__start.fingers = [newnode]*len(self.__start.fingers)
self.__start.initialized = True
return
predecessor = dyschord.find_predecessor(self.__start, newnode.id)
successor = predecessor.next
if successor.id == newnode.id :
raise Exception("Node already exists with same id")
successor.prepend_node(newnode)
# Optimize fingers. Don't need to lock the nodes while this is
# being done.
newnode.update_fingers()
dyschord.announce(newnode)
def leave(self, node) :
if self.num_nodes() == 0 :
return
# Note I am looking up, because then I can get a node to leave by
# passing in another instance with the same id. Might be useful
# for testing.
predecessor = dyschord.find_predecessor(self.__start, node.id)
# print "Leaving node %d has predecessor %d" % (node.id, predecessor.id)
if predecessor.next.id != node.id :
# No joined node with this id. Maybe log the missing node, but
# work is done
return
leaving = predecessor.next
# Check we aren't removing first item
if leaving.id == self.__start.id :
self.__start = leaving.next
leaving.leave()
