Java套接字编程——连接完成之前套接字超时

Question

我有一个套接字编程代码......

Server Program

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

public class GreetingServer extends Thread
{
private static int port;
  private ServerSocket serverSocket;

  public GreetingServer(int port) throws IOException
   {
  serverSocket = new ServerSocket(port);
  serverSocket.setSoTimeout(50000);
   }

 public void run()
  {
     while(true)
     {
        try
         {
           System.out.println("Waiting for client on port " +
           serverSocket.getLocalPort() + "...");
           Socket server = serverSocket.accept();
           System.out.println("Just connected to "
                + server.getRemoteSocketAddress());
           DataInputStream in =
              new DataInputStream(server.getInputStream());
           System.out.println(in.readUTF());
           DataOutputStream out =
             new DataOutputStream(server.getOutputStream());
            out.writeUTF("Thank you for connecting to "
             + server.getLocalSocketAddress() + "\nGoodbye!");
        server.close();
     }catch(SocketTimeoutException s)
     {
        System.out.println("Socket timed out!");
        break;
     }catch(IOException e)
     {
        e.printStackTrace();
        break;
       }
    }
 }
     public static void main(String [] args)
     {
       port=9000;
       try
     {
       Thread t = new GreetingServer(port);
       t.start();
     }catch(IOException e)
    {
      e.printStackTrace();
    }
   }
 }

客户端代码是...

import java.io.*;
import java.net.*;
public class GreetingClient
 {
   private static String serverName;
   public static void main(String [] args)
    {
      String sName = "MyServerName";
      int port = 9000;
      try
       {
        System.out.println("Connecting to " + sName
                         + " on port " + port);
     Socket client = new Socket(sName, port);
     System.out.println("Just connected to "
                  + client.getRemoteSocketAddress());
     OutputStream outToServer = client.getOutputStream();
     DataOutputStream out =
                   new DataOutputStream(outToServer);

     out.writeUTF("Hello from "
                  + client.getLocalSocketAddress());
     InputStream inFromServer = client.getInputStream();
     DataInputStream in =
                    new DataInputStream(inFromServer);
     System.out.println("Server says " + in.readUTF());
     client.close();
  }catch(IOException e)
    {
    }
  }
 }

代码编译良好。但是当我运行该程序时，首先是服务器，然后是客户端，服务器显示

Waiting for client on port 9000...
Socket timed out!

客户端显示

Connecting to MyServerName on port 9000

代码有什么问题吗？我尝试增加和减少超时值，但它给出相同的输出。

Answer 1

What is wrong with the code?

你的代码没有任何问题。您设置了 50 秒的接受超时，接受阻塞了 50 秒而没有客户端尝试连接，因此抛出了 SocketTimeoutException。这里的一切都按设计运行。

当然有可能:

• 您的接受超时时间太短
• 您不想仅仅因为接受超时而中止服务器
• 您根本不希望接受超时。

所有这些都是取决于您的要求的设计决策。

也有可能您在客户端中输入了错误的主机名，但由于您忽略了客户端中的异常:

catch(IOException e)
{
}

你永远不会知道。永远不要忽略IOExceptions。