Need help on chat between two computers

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.

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 – ∞)

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

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 – ∞)

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

larsp777

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

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

larsp777

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.

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

OriginalGriff

: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 – ∞)

larsp777

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

larsp777

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?

Rahul VB

hey OG, please enlighten me more on this.

OriginalGriff

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 – ∞)

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 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

larsp777

Thanks! It will probebly be on the same LAN.