I have a simple example of a socket server where the user connects to it and may or may not immediately send input (in this case it does not). Java throws an exception and I don't understand why.

Here's the code:

```
public static void main(String[] args) {
    org.apache.log4j.PropertyConfigurator.configure("conf/log4j.properties");
    log.debug("Application started");

    try {
        // Setup socket server
        int port = 9999;
        sv = new ServerSocket(port);

        // Accept incoming sockets or block
        Socket s = sv.accept();
        log.warn("Accepted new client connection!");

        InputStreamReader reader = new InputStreamReader(s.getInputStream(), "UTF-8");

        while(!reader.ready()) {}

        char[] b = null;
        reader.read(b);
        log.warn(new String(b));

    } catch (Exception e) {
        System.out.print(e.getMessage() + "\n");
        e.printStackTrace();
        System.exit(1);
    }
}

```

And the stack trace: 

```
DEBUG [main] (Main.java:38) - Application started
WARN [main] (Main.java:47) - Accepted new client connection!
null
java.lang.NullPointerException
    at java.io.Reader.read(Reader.java:140)
    at com.cinefyapp.Main.main(Main.java:51)

```

Thanks for helping.