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  4. Need help on chat between two computers

Need help on chat between two computers

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  • L larsp777

    Not sure I understand...Could you show how to change my code? This is my code:

    private void Form1_Load(object sender, EventArgs e)
    {
    //Set up socket
    sck = new Socket(AddressFamily.InterNetwork, SocketType.Dgram, ProtocolType.Udp);
    sck.SetSocketOption(SocketOptionLevel.Socket, SocketOptionName.ReuseAddress, true);

            //Get user IP
            txtLocalIP.Text = GetLocalIP();
            txtRemoteIP.Text = GetLocalIP();
        }
    
    
        private string GetLocalIP()
        {
            IPHostEntry host;
            host = Dns.GetHostEntry(Dns.GetHostName());
            foreach (IPAddress ip in host.AddressList)
            {
                if (ip.AddressFamily == AddressFamily.InterNetwork)
                    return ip.ToString();
            }
            return "127.0.0.1";
        }
    
        private void MessageCallBack(IAsyncResult aResult)
        {
            try
            {
                byte\[\] receivedData = new byte\[1500\];
                receivedData = (byte\[\])aResult.AsyncState;
    
                //Converting byte{\] to string
                ASCIIEncoding aEncoding = new ASCIIEncoding();
                string receivedMessage = aEncoding.GetString(receivedData);
    
                //Adding this messegage into ListBox
                listMessage.Items.Add("Friend: " + receivedMessage);
    
                buffer = new byte\[1500\];
                sck.BeginReceiveFrom(buffer, 0, buffer.Length, SocketFlags.None, ref epRemote, new AsyncCallback(MessageCallBack), buffer);
            }
            catch (Exception ex)
            {
                MessageBox.Show(ex.ToString());
            }
        }
    
    
        private void btnConnect\_Click(object sender, EventArgs e)
        {
            //Binding socket
            epLocal = new IPEndPoint(IPAddress.Parse(txtLocalIP.Text), Convert.ToInt32(txtLocalPort.Text));
            sck.Bind(epLocal);
    
            //Connecting to remote IP
            epRemote = new IPEndPoint(IPAddress.Parse(txtRemoteIP.Text), Convert.ToInt32(txtRemotePort.Text));
            sck.Connect(epRemote);
    
            //Listen to specific port
            buffer = new byte\[1500\];
            sck.BeginReceiveFrom(buffer, 0, buffer.Length, SocketFlags.None, ref epRemote, new AsyncCallback(MessageCallBack), buffer);
        }
    
        private void btnSend\_Click(object sender, EventArgs e)
        {
            //Convert string message to
    
    OriginalGriffO Offline
    OriginalGriffO Offline
    OriginalGriff
    wrote on last edited by
    #4

    Where do you get the error? Which line?

    Those who fail to learn history are doomed to repeat it. --- George Santayana (December 16, 1863 – September 26, 1952) Those who fail to clear history are doomed to explain it. --- OriginalGriff (February 24, 1959 – ∞)

    "I have no idea what I did, but I'm taking full credit for it." - ThisOldTony
    "Common sense is so rare these days, it should be classified as a super power" - Random T-shirt

    L 2 Replies Last reply
    0
    • L larsp777

      Not sure I understand...Could you show how to change my code? This is my code:

      private void Form1_Load(object sender, EventArgs e)
      {
      //Set up socket
      sck = new Socket(AddressFamily.InterNetwork, SocketType.Dgram, ProtocolType.Udp);
      sck.SetSocketOption(SocketOptionLevel.Socket, SocketOptionName.ReuseAddress, true);

              //Get user IP
              txtLocalIP.Text = GetLocalIP();
              txtRemoteIP.Text = GetLocalIP();
          }
      
      
          private string GetLocalIP()
          {
              IPHostEntry host;
              host = Dns.GetHostEntry(Dns.GetHostName());
              foreach (IPAddress ip in host.AddressList)
              {
                  if (ip.AddressFamily == AddressFamily.InterNetwork)
                      return ip.ToString();
              }
              return "127.0.0.1";
          }
      
          private void MessageCallBack(IAsyncResult aResult)
          {
              try
              {
                  byte\[\] receivedData = new byte\[1500\];
                  receivedData = (byte\[\])aResult.AsyncState;
      
                  //Converting byte{\] to string
                  ASCIIEncoding aEncoding = new ASCIIEncoding();
                  string receivedMessage = aEncoding.GetString(receivedData);
      
                  //Adding this messegage into ListBox
                  listMessage.Items.Add("Friend: " + receivedMessage);
      
                  buffer = new byte\[1500\];
                  sck.BeginReceiveFrom(buffer, 0, buffer.Length, SocketFlags.None, ref epRemote, new AsyncCallback(MessageCallBack), buffer);
              }
              catch (Exception ex)
              {
                  MessageBox.Show(ex.ToString());
              }
          }
      
      
          private void btnConnect\_Click(object sender, EventArgs e)
          {
              //Binding socket
              epLocal = new IPEndPoint(IPAddress.Parse(txtLocalIP.Text), Convert.ToInt32(txtLocalPort.Text));
              sck.Bind(epLocal);
      
              //Connecting to remote IP
              epRemote = new IPEndPoint(IPAddress.Parse(txtRemoteIP.Text), Convert.ToInt32(txtRemotePort.Text));
              sck.Connect(epRemote);
      
              //Listen to specific port
              buffer = new byte\[1500\];
              sck.BeginReceiveFrom(buffer, 0, buffer.Length, SocketFlags.None, ref epRemote, new AsyncCallback(MessageCallBack), buffer);
          }
      
          private void btnSend\_Click(object sender, EventArgs e)
          {
              //Convert string message to
      
      Richard DeemingR Offline
      Richard DeemingR Offline
      Richard Deeming
      wrote on last edited by
      #5

      larsp777 wrote:

      listMessage.Items.Add("Friend: " + receivedMessage);

      I'm guessing that's the line that's causing the problem. The callback method will be called on a background thread; you will need to use InvokeRequired and Invoke to get back to the UI thread when you want to access the form's controls.

      Invoke((Func<object, int>)listMessage.Items.Add, "Friend: " + receivedMessage);


      "These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer

      "These people looked deep within my soul and assigned me a number based on the order in which I joined" - Homer

      L 1 Reply Last reply
      0
      • Richard DeemingR Richard Deeming

        larsp777 wrote:

        listMessage.Items.Add("Friend: " + receivedMessage);

        I'm guessing that's the line that's causing the problem. The callback method will be called on a background thread; you will need to use InvokeRequired and Invoke to get back to the UI thread when you want to access the form's controls.

        Invoke((Func<object, int>)listMessage.Items.Add, "Friend: " + receivedMessage);


        "These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer

        L Offline
        L Offline
        larsp777
        wrote on last edited by
        #6

        Thank you both! I think that is the problem. Will check it out and get back.

        1 Reply Last reply
        0
        • OriginalGriffO OriginalGriff

          Where do you get the error? Which line?

          Those who fail to learn history are doomed to repeat it. --- George Santayana (December 16, 1863 – September 26, 1952) Those who fail to clear history are doomed to explain it. --- OriginalGriff (February 24, 1959 – ∞)

          L Offline
          L Offline
          larsp777
          wrote on last edited by
          #7

          Ok... The other guy whose thread I can´t se now wrote: I'm guessing that's the line that's causing the problem. The callback method will be called on a background thread; you will need to use InvokeRequired and Invoke to get back to the UI thread when you want to access the form's controls. Invoke((Action<object>)listMessage.Items.Add, "Friend: " + receivedMessage); I Think he was right about the line. Tried to do change the line accordingly but got an error:

          int System.Windows.Forms.ListBox.ObjectCollection.Add(object)' has the wrong return type

          Richard DeemingR 1 Reply Last reply
          0
          • OriginalGriffO OriginalGriff

            Where do you get the error? Which line?

            Those who fail to learn history are doomed to repeat it. --- George Santayana (December 16, 1863 – September 26, 1952) Those who fail to clear history are doomed to explain it. --- OriginalGriff (February 24, 1959 – ∞)

            L Offline
            L Offline
            larsp777
            wrote on last edited by
            #8

            The error occurs when I try to send a message. It occurs on both computers. When I send from one computer the error shows on the other.

            1 Reply Last reply
            0
            • L larsp777

              Ok... The other guy whose thread I can´t se now wrote: I'm guessing that's the line that's causing the problem. The callback method will be called on a background thread; you will need to use InvokeRequired and Invoke to get back to the UI thread when you want to access the form's controls. Invoke((Action<object>)listMessage.Items.Add, "Friend: " + receivedMessage); I Think he was right about the line. Tried to do change the line accordingly but got an error:

              int System.Windows.Forms.ListBox.ObjectCollection.Add(object)' has the wrong return type

              Richard DeemingR Offline
              Richard DeemingR Offline
              Richard Deeming
              wrote on last edited by
              #9

              larsp777 wrote:

              int System.Windows.Forms.ListBox.ObjectCollection.Add(object)' has the wrong return type

              Whoops - didn't notice that the method returned an int. The delegate type will need to be a Func<object, int> instead of an Action<object>.

              Invoke((Func<object, int>)listMessage.Items.Add, "Friend: " + receivedMessage);


              "These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer

              "These people looked deep within my soul and assigned me a number based on the order in which I joined" - Homer

              OriginalGriffO L 3 Replies Last reply
              0
              • Richard DeemingR Richard Deeming

                larsp777 wrote:

                int System.Windows.Forms.ListBox.ObjectCollection.Add(object)' has the wrong return type

                Whoops - didn't notice that the method returned an int. The delegate type will need to be a Func<object, int> instead of an Action<object>.

                Invoke((Func<object, int>)listMessage.Items.Add, "Friend: " + receivedMessage);


                "These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer

                OriginalGriffO Offline
                OriginalGriffO Offline
                OriginalGriff
                wrote on last edited by
                #10

                :thumbsup:

                Those who fail to learn history are doomed to repeat it. --- George Santayana (December 16, 1863 – September 26, 1952) Those who fail to clear history are doomed to explain it. --- OriginalGriff (February 24, 1959 – ∞)

                "I have no idea what I did, but I'm taking full credit for it." - ThisOldTony
                "Common sense is so rare these days, it should be classified as a super power" - Random T-shirt

                1 Reply Last reply
                0
                • Richard DeemingR Richard Deeming

                  larsp777 wrote:

                  int System.Windows.Forms.ListBox.ObjectCollection.Add(object)' has the wrong return type

                  Whoops - didn't notice that the method returned an int. The delegate type will need to be a Func<object, int> instead of an Action<object>.

                  Invoke((Func<object, int>)listMessage.Items.Add, "Friend: " + receivedMessage);


                  "These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer

                  L Offline
                  L Offline
                  larsp777
                  wrote on last edited by
                  #11

                  Thanks, Richard, it worked! Well, thank you both for helping me out!

                  1 Reply Last reply
                  0
                  • L larsp777

                    I have used this tutorial to make a chat: http://www.youtube.com/watch?v=BDVfpPq3weo[^] It works on the same computer but not between two. The problem seem so be that I need more than one thread. I get Cross-thread operation not valid.

                    L Offline
                    L Offline
                    larsp777
                    wrote on last edited by
                    #12

                    Can you mark a question as solved. You really solved it Richard but Griffin also helped. I am working on a game for XNA and need both computers to have access to the same "board". Do you know anything about that?

                    1 Reply Last reply
                    0
                    • OriginalGriffO OriginalGriff

                      No, the problem is the reverse of that: "Cross-thread operation not valid" means that you are executing code on one thread that can only be executed on a different thread: normally, this occurs when you try to update a control from a different thread to that from which it was created (which must be the UI thread), either in a BackgoundWorker, a Thread instance or in the handler of a communications class that uses threading to handle it's events (the SerialPort does this for example). Check your code: you may just have to start invoking the control instead of accessing it directly. For example:

                          private void AddNewTab(string tabName)
                              {
                              if (InvokeRequired)
                                  {
                                  Invoke(new MethodInvoker(delegate { AddNewTab(tabName); }));
                                  }
                              else
                                  {
                                  TabPage tp = new TabPage(tabName);
                                  myTabControl.TabPages.Add(tp);
                                  }
                              }
                      

                      Those who fail to learn history are doomed to repeat it. --- George Santayana (December 16, 1863 – September 26, 1952) Those who fail to clear history are doomed to explain it. --- OriginalGriff (February 24, 1959 – ∞)

                      R Offline
                      R Offline
                      Rahul VB
                      wrote on last edited by
                      #13

                      hey OG, please enlighten me more on this.

                      OriginalGriffO 1 Reply Last reply
                      0
                      • R Rahul VB

                        hey OG, please enlighten me more on this.

                        OriginalGriffO Offline
                        OriginalGriffO Offline
                        OriginalGriff
                        wrote on last edited by
                        #14

                        Sorry Rahul, but it's a huge subject - I couldn't begin to do it justice in a small text box! :laugh: There are some good tutorials out there which explain threading and the UI pretty well: http://stuff.seans.com/2009/05/21/net-basics-do-work-in-background-thread-to-keep-gui-responsive/[^] (Backgound and why to thread) http://www.dreamincode.net/forums/topic/246911-c%23-multi-threading-in-a-gui-environment/[^] (Fairly advanced) But basically when you start to use multiple threads you can't touch any controls, except from the thread that created them - which is called the UI thread (for User Interface) and is the original thread the form started on. If you try, you will get a "cross-threading" error telling you not to do that. The only way to get round it is to Invoke the control - which basically requests the UI thread to do the work for you. Have a look at the BackgroundWorker thread - it provides a way to update the display without invoking via the ProgressChanged event, which is executed on the original thread.

                        Those who fail to learn history are doomed to repeat it. --- George Santayana (December 16, 1863 – September 26, 1952) Those who fail to clear history are doomed to explain it. --- OriginalGriff (February 24, 1959 – ∞)

                        "I have no idea what I did, but I'm taking full credit for it." - ThisOldTony
                        "Common sense is so rare these days, it should be classified as a super power" - Random T-shirt

                        1 Reply Last reply
                        0
                        • Richard DeemingR Richard Deeming

                          larsp777 wrote:

                          int System.Windows.Forms.ListBox.ObjectCollection.Add(object)' has the wrong return type

                          Whoops - didn't notice that the method returned an int. The delegate type will need to be a Func<object, int> instead of an Action<object>.

                          Invoke((Func<object, int>)listMessage.Items.Add, "Friend: " + receivedMessage);


                          "These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer

                          L Offline
                          L Offline
                          larsp777
                          wrote on last edited by
                          #15

                          Do you know if there is a simple way for a third person can listen in what the others are Writing. (only has to listen to the sender) I am making an application where you should be able to simulate someone eavesdropping the conversation.

                          Richard DeemingR 1 Reply Last reply
                          0
                          • L larsp777

                            Do you know if there is a simple way for a third person can listen in what the others are Writing. (only has to listen to the sender) I am making an application where you should be able to simulate someone eavesdropping the conversation.

                            Richard DeemingR Offline
                            Richard DeemingR Offline
                            Richard Deeming
                            wrote on last edited by
                            #16

                            If all the computers are on the same LAN, you could use the broadcast address so that any computer can pick up the messages: Broadcasting Using Socket-Oriented Approach[^] If they're on different networks, or you don't want the overhead associated with broadcasting, then you'll need to use multicasting: IP Multicasting in C#[^]


                            "These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer

                            "These people looked deep within my soul and assigned me a number based on the order in which I joined" - Homer

                            L 1 Reply Last reply
                            0
                            • Richard DeemingR Richard Deeming

                              If all the computers are on the same LAN, you could use the broadcast address so that any computer can pick up the messages: Broadcasting Using Socket-Oriented Approach[^] If they're on different networks, or you don't want the overhead associated with broadcasting, then you'll need to use multicasting: IP Multicasting in C#[^]


                              "These people looked deep within my soul and assigned me a number based on the order in which I joined." - Homer

                              L Offline
                              L Offline
                              larsp777
                              wrote on last edited by
                              #17

                              Thanks! It will probebly be on the same LAN.

                              1 Reply Last reply
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