A while ago I developed a little LAN chat app. in Java which allows chatting with other hosts, send images, etc. Although it was created just for fun, now it's being used where I work.

Currently, there is no "chat server" on the app. where each client registers, updates it's status, etc. (I liked the idea of symmetric design and not depending on a server running on some other machine).

Instead, each host is a client/server which has a hosts.properties file with the hostname of the other hosts, and - for instance - broadcasts to each one of them when sending a massive message/image/whatever.

In the beginning there were just a couple of hosts, so this hosts.properties file wasn't an issue. But as the amount of users increased, the need of updating that file was a bit daunting. So now I've decided to get rid of it, and each time the app. starts, dynammically find the other active hosts.

However, I cannot find the correct way of implement this. I've tried starting different threads, each one of them searching for other hosts in a known range of IP addresses. Something like this (simplified for the sake of readability):

/** HostsLocator */
public static void searchForHosts(boolean waitToEnd) {
    for (int i=0; i < MAX_IP; i+= MAX_IP / threads) {
        HostsLocator detector = new HostsLocator(i, i+(MAX_IP / threads - 1)); // range: from - to
        new Thread(detector).start();                 
    }
}

public void run() {
    for (int i=from; i<=to; i++)
        findHosts( maskAddress + Integer.toString(i) );
}

public static boolean findHosts(String IP) {
    InetAddress address = InetAddress.getByName(IP);
    if ( address.isReachable(CONNECTION_TIME_OUT) )
        // host found!
}

However:

• With a single thread and a low value in CONNECTION_TIME_OUT (500ms) I get wrong Host Not Found status for for hosts actually active.
• With a high value in CONNECTION_TIME_OUT (5000ms) and only one single thread takes forever to end
• With several threads I've also found problems similar like the first one, due to collisions.

So... I guess there's a better way of solving this problem but I couldn't find it. Any advice? Thanks!