由网友(Airport（空港）)分享简介：我是 Java 编程新手，我正在尝试创建一个 UDP 服务器.当我编译代码时，它说它无法监听端口 4722，我想知道为什么.下面是代码.如有任何建议，我将不胜感激.I am new to Java programming and I am trying to create a UDP server. When I c...

我是 Java 编程新手，我正在尝试创建一个 UDP 服务器.当我编译代码时，它说它无法监听端口 4722，我想知道为什么.下面是代码.如有任何建议，我将不胜感激.

I am new to Java programming and I am trying to create a UDP server. When I compile the code it says it could not listen to port 4722 and I would like to know why. Below is the code. I would be grateful for any advice.

import java.net.*;
import java.io.*;


public class Server 
{
public static void main(String[] args) throws IOException 
{
    DatagramSocket serverSocket = new DatagramSocket(4722);
    Socket clientSocket = null;
    byte[] receiveData = new byte[1024];
    byte[] sendData = new byte [1024];      

    boolean command = true;
        try
        {
            serverSocket = new DatagramSocket(4722);
            DatagramPacket receivePacket = new DatagramPacket(receiveData,receiveData.length);
            System.out.println("Waiting for client...");


        } 


        catch (IOException e) 
        {
            System.err.println("Could not listen on port: 4722.");
            System.exit(1);

        }
        DatagramPacket packet = new DatagramPacket (sendData,sendData.length,4722);
        serverSocket.send(packet);



        PrintWriter out = new PrintWriter(clientSocket.getOutputStream(), true);
        BufferedReader in = new BufferedReader(new InputStreamReader(clientSocket.getInputStream()));
        String inputLine, outputLine;
        mathematicalProtocol bbm = new mathematicalProtocol();

        outputLine = bbm.processInput(null);
        out.println(outputLine);

        while ((inputLine = in.readLine()) != null) 
        {
            if(inputLine.equals("Bye."))
            break;
            outputLine = bbm.processInput(inputLine);
            out.println(outputLine);
            if (outputLine.equals("Bye."))

            break;
        }

        out.close();
        in.close();
        clientSocket.close();
        serverSocket.close();
}
}

推荐答案

你正在初始化 serverSocket 然后再次在同一个端口上创建一个新的 DatagramSocket(你可以'不要这样做，因为它已经绑定在第一个 DatagramSocket 上).IE.删除以下行:

You are initializing serverSocket and then making a new DatagramSocket on the same port again (and you can't do that as it's already bound on the first DatagramSocket). I.e. remove the following line:

        serverSocket = new DatagramSocket(4722);