为啥服务器在非多线程的情况下会同时处理多个客户端？

Posted 2023-04-15

【中文标题】为啥服务器在非多线程的情况下会同时处理多个客户端？

【英文标题】：Why does server process multiple clients at the same time when it's not multithreaded?为什么服务器在非多线程的情况下会同时处理多个客户端？

【发布时间】：2021-05-17 17:30:12

【问题描述】：

所以在下面的代码中，我有一个单线程服务器和一个多线程客户端。这样做的原因是因为我希望客户端同时发送数据包和接收数据包。但是，当我启动服务器并运行多个客户端时，服务器可以同时处理多个客户端，即使服务器不是多线程的？你能解释一下吗？

服务器：

public class server 

public static void main(String[] args) throws IOException
    new server();



//declare the UDP server socket
DatagramSocket datagramSocket;
int port = 3741;

public server() throws IOException 
    
    //create UDP server with a specific port number
    datagramSocket = new DatagramSocket(port);
    System.out.println("[UDP server] Datagram socket started on port " + port);
    
    //prepare the packet structure for received packets
    int dataLength = 100; //must be large enough so some part of packet doesn't get lost
    byte[] receiveData = new byte[dataLength];
    DatagramPacket packet = new DatagramPacket(receiveData, receiveData.length);
    
while(true) 
        datagramSocket.receive(packet);
        System.out.println("client connected");
        
        InetAddress inetAddress = packet.getAddress();
        int clientPort = packet.getPort();
        byte[] data = packet.getData();

        
        
        //send response back to the client host
        byte[] sendData = new byte[1024];
        
        DatagramPacket datagramPacket = new DatagramPacket(data, data.length, inetAddress, clientPort);
        datagramSocket.send(datagramPacket); //sending data from server to client
        

客户：

public class client 

public static void main(String[] args) throws Exception 
    new client();



public client() throws Exception
    
    DatagramSocket socket = new DatagramSocket();
    
    InetAddress ip = InetAddress.getByName("127.0.0.1");
    String message = "hello from client"; 
    
    DatagramPacket packetSend = new DatagramPacket(message.getBytes(), message.getBytes().length, ip, 3741);
    
    Thread th = new Thread(new Runnable() 
        
        @Override
        public void run() 
            for(int i = 0; i < 100; i++) 
                try 
                    Thread.sleep(1000);
                 catch (InterruptedException e) 
                    // TODO Auto-generated catch block
                    e.printStackTrace();
                
                
                try 
                    socket.send(packetSend);
                 catch (IOException e) 
                    // TODO Auto-generated catch block
                    e.printStackTrace();
                
            
            
        
    );
    th.start();
    
    byte[] buffer = new byte[1024];
    DatagramPacket packet = new DatagramPacket(buffer, buffer.length);
    
    //this part can't be in a thread because the loop above us will finish first before this starts
    //we can put this code before the loop and start a thread this would also work
    while(true) 
        try 
            socket.receive(packet);
         catch (IOException e) 
            // TODO Auto-generated catch block
            e.printStackTrace();
        
        
        String data = new String(packet.getData()).trim();
        
        System.out.println(data);
    

【问题讨论】：

“服务器可以同时处理多个客户端”-请编辑问题更具体-您到底看到了什么对您没有意义？来自不同客户端的交错打印，每个请求一个？

【参考方案1】：

你认为 “服务器可以同时处理多个客户端，即使服务器不是多线程的” 会导致你的大脑欺骗你。实际上，他们正在以相对较快的速度逐一处理。很容易看出您是使用wireshark 还是tcpdump 来捕获服务器端数据包。你会发现有趣的真相。

【讨论】：

我将 thread.sleep() 放在服务器中，它显示了每个客户端是如何被一一处理的。无需为此用途安装应用程序。这样可以更轻松地完成。

如果想深入网络编程，总有一天需要wireshark和tcpdump。

我是一名全栈 Web 开发人员，没有必要深入研究任何特定主题。