将VM Client连接到我的计算机上的服务器C#

时间:2020-05-31 22:51:10

标签: c# server client tcplistener

我遇到的此网络聊天应用程序代码遇到了麻烦。当两个应用程序都在同一台计算机上运行时,它运行良好,但是当尝试将VM客户端连接到在实际安装中运行的主机应用程序时,出现超时。这是代码。

static readonly object _lock = new object();
        static readonly Dictionary<int, TcpClient> list_clients = new Dictionary<int, TcpClient>();

        static void Main(string[] args) {
            int count = 1;

            TcpListener ServerSocket = new TcpListener(IPAddress.Parse("192.168.1.59"), 5000);
            ServerSocket.Start();

            while(true) {
                TcpClient client = ServerSocket.AcceptTcpClient();
                lock(_lock) list_clients.Add(count, client);
                Console.WriteLine("Someone connected!!");

                Thread t = new Thread(handle_clients);
                t.Start(count);
                count++;
            }
        }
    }

class Client {
        static void Main(string[] args) {
            IPAddress ip = IPAddress.Parse("192.168.1.59");
            int port = 5000;
            TcpClient client = new TcpClient();
            client.Connect(ip, port);
            Console.WriteLine("client connected!!");
            NetworkStream ns = client.GetStream();
            Thread thread = new Thread(o => ReceiveData((TcpClient)o));

            thread.Start(client);

            string s;
            while(!string.IsNullOrEmpty((s = Console.ReadLine()))) {
                byte[] buffer = Encoding.ASCII.GetBytes(s);
                ns.Write(buffer, 0, buffer.Length);
            }

            client.Client.Shutdown(SocketShutdown.Send);
            thread.Join();
            ns.Close();
            client.Close();
            Console.WriteLine("disconnect from server!!");
            Console.ReadKey();
        }
    }```

1 个答案:

答案 0 :(得分:1)

您的VM可能正在不同的网络接口中运行。

请尝试:

TcpListener ServerSocket = new TcpListener(IPAddress.Any, 5000);

通过这种方式,您可以告诉服务器在任何网络接口上进行监听。

添加:此外,VM客户端应连接到计算机(使用TCPListener程序)在VM网络范围内的IP地址。不是192.168.1.59,而是:

static void Main(string[] args) {
    IPAddress ip = IPAddress.Parse("172.17.xx.xx");
    ...