Java服务器套接字

Question

我遇到了一个小问题。我想使用服务器套接字将玩家列表发送到客户端。但由于某种原因，当我尝试运行应用程序时，它会在到达 socket = serverSocket.accept(); 时停止。我在谷歌上尝试了一些东西，但不起作用。

package Serverside;
import java.io.*;
import java.net.*;
import com.mygdx.game.Sprites.Hero;
import com.mygdx.game.Sprites.Player;
import java.util.ArrayList;
/**
 *
 * @author Tjidde Nieuwenhuizen
 */
public class ServerArenaOfLegends {
    static ServerSocket serverSocket;
    static Socket socket;
    static ObjectOutputStream outStreamObj;
    static ObjectInputStream inStreamObj;
    static ArrayList<Player> playerList;
    public static void main(String[] args) {
        ServerArenaOfLegends sr = new ServerArenaOfLegends();
        sr.run();
    }
    private void run() {
        playerList = new ArrayList<Player>();
        Player p1;
        Hero hero = new Hero(2, null, 3);
        p1 = new Player(null, null, hero);
        playerList.add(p1);
        try {
            serverSocket = new ServerSocket(5555);
            socket = serverSocket.accept();
            outStreamObj = new ObjectOutputStream(socket.getOutputStream());
            while (true) {
                outStreamObj.writeObject(playerList);
            }
        } catch (Exception ex) {
            System.out.println(ex.toString());
        }
    }
}

Answer 1

如果您运行此示例:

public static void main(String[] args) {

        try {
            ServerSocket serverSocket = new ServerSocket(5555);
            Socket socket = serverSocket.accept();
            System.out.println("DONE");
        } catch (IOException ex) {
            System.out.println(ex.getMessage());
        }
    }

程序将显示为挂起。

然后在命令行上运行“telnet 127.0.0.1 5555” 然后程序将打印“DONE”，然后结束。

所以.accept()阻塞直到它获取数据，这就是您所看到的行为。