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数据报套接字上的setSotimeout

[英]setSotimeout on a datagram socket

The server acts like an echo server. 服务器就像一个echo服务器。 The clients sends 10 packets to server (1 sec of gap) 客户端向服务器发送10个数据包(1秒的间隙)

When Client receives packets from the server, sometimes the packets are lost. 当客户端从服务器接收数据包时,有时数据包会丢失。

So the client has to wait for up to one second for the packet to arrive. 因此,客户端必须等待最多一秒才能使数据包到达。 If the packet does not arrive in 1 second then the client should continue sending the other packets. 如果数据包未在1秒内到达,则客户端应继续发送其他数据包。

How would i use .setSoTimeout to achieve this? 我将如何使用.setSoTimeout来实现这一目标?

Code: 码:

import java.io.*;
import java.net.*;
import java.util.*;
/*
* Client to process ping requests over UDP.
*/
public class PingClient
{
    private static final int AVERAGE_DELAY = 100; // milliseconds
    public static void main(String[] args) throws Exception
    {
// Get command line argument.
        int port = Integer.parseInt(args[1]);//specified as argument
// Create random number generator for use in simulating
// packet loss and network delay.
        System.out.println("Port "+port);
// Create a datagram socket for receiving and sending UDP packets
// through the port specified on the command line.
        DatagramSocket socket = new DatagramSocket(1234);

    int i=0;
        for(i=0;i<10;i++)
    {
    byte[] buf = new byte[1024] ;
    Calendar cal=Calendar.getInstance();
    String ping="Ping "+ i +" "+cal.getTimeInMillis()+"\r\n";
    buf=ping.getBytes("UTF-8");
    InetAddress address = InetAddress.getByName(args[0]);
    System.out.println("Name "+args[1]);
    DatagramPacket packet = new DatagramPacket(buf, buf.length, 
                                       address, port);
    packet.setData(buf);
    socket.send(packet);
    Thread.sleep( 10* AVERAGE_DELAY);//1 sec

    DatagramPacket server_response = new DatagramPacket(new byte[1024], 1024);
    // Block until the host receives a UDP packet.

        socket.setSoTimeout(1000); //I don't know how to use this
        socket.receive(server_response);

    // Print the recieved data.

        printData(server_response);

}   
}

private static void printData(DatagramPacket request) throws Exception
    {
// Obtain references to the packet's array of bytes.
    byte[] buf = request.getData();
// Wrap the bytes in a byte array input stream,
// so that you can read the data as a stream of bytes.
    ByteArrayInputStream bais = new ByteArrayInputStream(buf);
// Wrap the byte array output stream in an input stream reader,
// so you can read the data as a stream of characters.
    InputStreamReader isr = new InputStreamReader(bais);
// Wrap the input stream reader in a bufferred reader,
// so you can read the character data a line at a time.
// (A line is a sequence of chars terminated by any combination of \r and \n.)
    BufferedReader br = new BufferedReader(isr);
// The message data is contained in a single line, so read this line.
    String line = br.readLine();
// Print host address and data received from it.
    System.out.println(
        "Received from " +
        request.getAddress().getHostAddress() +
        ": " +
        new String(line) );
    }

} }

The javadoc for setSoTimeout says: setSoTimeoutjavadoc说:

With this option set to a non-zero timeout, a call to receive() for this DatagramSocket will block for only this amount of time. 如果将此选项设置为非零超时,则对此DatagramSocket的receive()调用将仅阻止这段时间。 If the timeout expires, a java.net.SocketTimeoutException is raised, though the DatagramSocket is still valid. 如果超时到期,则引发java.net.SocketTimeoutException,尽管DatagramSocket仍然有效。

So, if you want to send packets if no response has been received after 1 second, you just have to use 因此,如果您想在1秒后未收到任何响应时发送数据包,则只需使用即可

socket.setSoTimeout(1000L);
boolean continueSending = true;
int counter = 0;
while (continueSending && counter < 10) {
    // send to server omitted
    counter++;
    try {
        socket.receive(packet);
        continueSending = false; // a packet has been received : stop sending
    }
    catch (SocketTimeoutException e) {
        // no response received after 1 second. continue sending
    }
}

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