Java socket bind twice without exception

Question

When my Application, it bind on a given port when it started:

public boolean checkIsAlreadyStart() {
    try {
        final ServerSocket server = new ServerSocket();
        server.setReuseAddress(false);
        server.bind(new InetSocketAddress("127.0.0.1",APPLICATION_PORT ));
        if (server.isBound()){
            logger.debug("binding to port: {}", APPLICATION_PORT);
            return false;
        }
        return true;
    } catch (IOException e) {
        logger.error("cannot bind to port", e);
        return true;
    }
}

However, when i run two instance of the Application at the same time, the second instance can still run the method without the IOException. Do I have to call accept() method?

user207421 · Accepted Answer · 2013-04-10 05:57:13Z

2

Cannot reproduce:

10/04/2013 3:55:34 PM S checkIsAlreadyStart
INFO: Bound to port: 9,999
10/04/2013 3:55:34 PM S checkIsAlreadyStart
SEVERE: cannot bind to port
java.net.BindException: Address already in use: JVM_Bind
    at java.net.PlainSocketImpl.socketBind(Native Method)
    at java.net.PlainSocketImpl.bind(PlainSocketImpl.java:383)
    at java.net.ServerSocket.bind(ServerSocket.java:328)
    at java.net.ServerSocket.bind(ServerSocket.java:286)
    at S.checkIsAlreadyStart(S.java:33)
    at S.main(S.java:63)

However as server is a local variable it is likely to allow garbage-collection of the ServerSocket, which will close it. If your intention is to keep the port open you should make server an instance variable.

answered Apr 10, 2013 at 5:57

user207421

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Comments

Beatty · Accepted Answer · 2013-04-10 02:33:00Z

0

Are you sure the first instance of the application is still running? You could place a Thread.sleep in the application to be sure it is still running when you launch the second instance.

I ran this twice and the second time failed, as you intended.

    public static void main(String args[]) {
    Menu m = new Menu();
    m.checkIsAlreadyStart();
    try {
    Thread.sleep(20000);
    } catch (Exception e){}
}

answered Apr 10, 2013 at 2:33

Beatty

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