c# - 客户端未连接到服务器，而是获得了无效的指针地址

Question

我想从我的电脑连接服务器。但它不起作用。它总是显示错误。我也想从客户端向服务器发送消息，所以请帮助我该怎么办？

The system detected an invalid pointer address in attempting to use a pointer argument in a call

我的代码是:

public partial class Form1 : Form
    {
        Socket socket;
        IPAddress ipAddress;
        string address = "192.168.1.203";
        public Form1()
        {
            InitializeComponent(); 
        }

        private void btnStartServer_Click(object sender, EventArgs e)
        {
            SocketPermission socketPermission = new SocketPermission(NetworkAccess.Connect, TransportType.Tcp, "", SocketPermission.AllPorts);
            socketPermission.Demand();
            IPHostEntry ipHOst = Dns.GetHostEntry("");
            ipAddress = ipHOst.AddressList[0];
            IPEndPoint ipEndPoint = new IPEndPoint(IPAddress.Parse(address), 4532);
            socket = new Socket(ipAddress.AddressFamily, SocketType.Stream, ProtocolType.Tcp);
            socket.NoDelay = false;
            socket.Connect(ipEndPoint);
            txtClinet1.Text = "Socket connected to " + socket.RemoteEndPoint.ToString();
        }

    }

Answer 1

是什么让您认为AddressList [0]是可以使用的地址列表？根据DOCS，您必须遍历该列表，直到找到正确的列表为止。
例如:

private void button1_Click(object sender, EventArgs e)
    {
        SocketPermission socketPermission = new SocketPermission(NetworkAccess.Connect, TransportType.Tcp, "", SocketPermission.AllPorts);
        socketPermission.Demand();
        IPHostEntry ipHOst = Dns.GetHostEntry("");
        for (int i = 0; i < ipHOst.AddressList.Length; i++)
        {

            ipAddress = ipHOst.AddressList[i];
            IPEndPoint ipEndPoint = new IPEndPoint(IPAddress.Parse(address), 4532);
            socket = new Socket(ipAddress.AddressFamily, SocketType.Stream, ProtocolType.Tcp);
            socket.NoDelay = false;
            try
            {
                socket.Connect(ipEndPoint);
                MessageBox.Show("Socket connected to " + socket.RemoteEndPoint.ToString());
                break;
            }
            catch (Exception eX)
            {
                MessageBox.Show(eX.Message);
            }
        }

    }

在我的计算机中，它连接到第三个地址 AddressList[2];