Java网络编程：关于入门socket的几个疑问

1.关闭输入输出流会自动关闭socket实例：关闭输出流之后再次获取输入流会抛错。

Socket socket = new Socket("127.0.0.1", 8001);
OutputStream outputStream = socket.getOutputStream();
DataOutputStream out = new DataOutputStream(outputStream);
out.writeUTF("Hello World");
out.flush();

outputStream.close();
System.out.println("socket是否关闭：" + socket.isClosed());// 输出true

InputStream inputStream = socket.getInputStream();

 

2.基于UDP连接demo：

问题：相同地址（127.0.0.1 8088）只能实例化一个DatagramSocket实例，服务器启动之后客户端启动会报地址占用错误！

代码如下：

服务端：

import java.io.IOException;
import java.net.DatagramPacket;
import java.net.DatagramSocket;
import java.net.SocketException;

/**
 * @Author: Lon
 * @Date: 2019/5/27 14:08
 * @Description:
 */
public class DatagramSocketServer {

    public static void main(String[] args) {
        try {
            System.out.println("启动服务，等待连接");
            byte[] bytes = new byte[1024];
            DatagramPacket packet = new DatagramPacket(bytes, bytes.length);

            DatagramSocket socket = new DatagramSocket(8088);
            socket.receive(packet);

            String string = new String(bytes, 0, bytes.length);
            System.out.println("收到消息：" + string);
        } catch (SocketException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}

客户端：

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

/**
 * @Author: Lon
 * @Date: 2019/5/27 13:59
 * @Description:
 */
public class DatagramSocketClient {

    public static void main(String[] args) {
        try {
            byte[] data = "Hello World...".getBytes();
            InetAddress address = InetAddress.getLocalHost();

            DatagramPacket packet = new DatagramPacket(data, data.length, address, 8088);
            DatagramSocket socket = new DatagramSocket();

            socket.send(packet);
            socket.close();
        } catch (UnknownHostException e) {
            e.printStackTrace();
        } catch (SocketException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}