Re: Peer-Discovery in own Group

I also suggest you use the rendezvous event notification callbacks and 
wait rather than sleep,  as it provides for a cleaner application 
interaction. Here's what I mean ;

private String rendezvousLock = "Rendezvous Lock";

..
..

rendezvous.addListener(this);

if ( !rendezvous.isRendezVous()  && !rendezvous.isConnectedToRendezVous()) {
            System.out.println("Waiting for Rendezvous Connection");
            try {
                synchronized(rendezvousLock) {
                    rendezvousLock.wait(autoRendezvousTimeout);
                }
            } catch (InterruptedException e) {
            }
        }
....
....

    public void rendezvousEvent(RendezvousEvent event) {
        if (event.getType() == event.RDVCONNECT ||
            event.getType() == event.RDVRECONNECT) {
            synchronized(rendezvousLock) {
                rendezvousLock.notify();
            }
        }
     }



Yun Ding wrote:

> Hi,
>
>> thank you very much for that explanation. Now I tried this:
>>
>> dhtPeerGoup.getRendezVousService().setAutoStart(true);
>> rdv = dhtPeerGoup.getRendezVousService();
>>  
>> while (!rdv.isConnectedToRendezVous()) {
>>    try {
>> Thread.sleep(2000);
>>    } catch (InterruptedException e1) {
>> }
>>
>> But I never get out of the loop. When there is just one instance of 
>> my Application. So there ist just one member in this group - me. So I 
>> should become rendezvous-peer of this group, don't I?
>>  
>>
> yes, you are right.  Your peer will become the rendezvous peer, and it 
> is not
> connected to itself (sounds strange).  You should use the following code.
> Your peer will get out of the loop, since it itself becomes a 
> rendezvous peer.
>
>        while (( ! rdv.isConnectedToRendezVous () ) &&
>           ( ! rdv.isRendezVous ()   ) )  {
>              try {
>                   Thread.sleep ( 2000  );
>              }
>          catch ( Exception e ) {}
>        }
>
> Best wishes,
>
> Yun
>
>


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