hi there, i create a tree there are 5 nodes and that are relevant each other.
ex:
Node 1 : childs : 2 and 3
Node 2: childs : 1 and 4
Node 3: childs 1 and 5
Node 4: childs 2
Node 5: childs 2
i created 5 thread at the same time.
foreach (Node dugum in nodes) { try { ParameterizedThreadStart ths = new ParameterizedThreadStart(StartTh); Thread th = new Thread(ths); th.Priority = ThreadPriority.Highest; th.Name = dugum.nodeid.ToString(); th.Start(dugum); } catch (ThreadStartException ex) { MessageBox.Show(ex.Message); } catch (ThreadInterruptedException ex) { MessageBox.Show(ex.Message); } }
void StartTh(object values) { Node node = (Node)values; foreach (Node child in node.childs) { listBox1.Items.Add("Nodeid :" + node.nodeid + " -> " + sendMesg(child)); }
}
private string sendMesg(Node child) {
string rslt= ""; lock (this) { try { if (!hashtable.ContainsKey(child.nodeid)) { hashtable.Add(child.nodeid, child.energy); Monitor.Wait(this); } Monitor.Pulse(this); hashtable.Remove(child.nodeid); rslt = child.nodeid.ToString(); } catch (ThreadStartException ex) { MessageBox.Show(ex.Message); } catch (ThreadInterruptedException ex) { MessageBox.Show(ex.Message); } } return rslt; }
my question is i don't want 2 nodes send same child at the same times. one node send and other wait in queue then send message. I created some codes but i cant see well solution.
the output always appears 5 or 6 result in listbox. actually i want to see 8 message to response about sending.
How can i do this with monitor ?
