?
Solved

What actually "label.visble = true" did?

Posted on 2004-10-13
3
Medium Priority
?
191 Views
Last Modified: 2010-04-15
The code below is extracted from one of my project, it will hang when there a client is connected. When I make some changes in the showConnected() function, the program work well. I would like to know why!!


//The code below is at the main page

private void btOpenPort_Click(object sender, System.EventArgs e)
{
      Connector = new TcpServer(Int32.Parse(tbPort.Text), this);
      Thread runSrv = new Thread(new ThreadStart(Connector.OpenPort));
      runSrv.Start();
}

public void showConnected()
{
                //lbConnected is a label
      lbConnected.visible = true;
}

//These the code run is another thread

public TcpServer(int port, Server theBoss)
{
      PortNo = port;
      boss = theBoss;
}

public void OpenPort()
{
      Listener = new TcpListener(PortNo);;
      Listener.Start();
                  
                  
      Socket ConnectionSocket;
      NetworkStream socketStream;
      BinaryReader reader;
      BinaryWriter writer;

      while(!programClosed)
      {
            ConnectionSocket = Listener.AcceptSocket();
                        
            boss.showConnected();
                        
            socketStream = new NetworkStream(ConnectionSocket);

            writer = new BinaryWriter(socketStream);
            reader = new BinaryReader(socketStream);
            ClientReader = reader;
      }
}


When I initially set the visible of lbConnected.visible = false and lbConnected.Text = "" and modified the showConnected function as below, the program work well, why?

public void showConnected()
{
      lbConnected.Text = "Client Connected !!";
}
0
Comment
Question by:Kahhoe
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 2
3 Comments
 
LVL 1

Accepted Solution

by:
CrazyIvan007 earned 500 total points
ID: 12306787
You are accessing the UI Thread from another Thread. That can makes problem. You should ever use Delegates to access the UI Thread from within Threads.


public delegate void DelegateSetLabelVisible(string strControlName, bool boolVisible);

public class YourClass
{
     public DelegateSetLabelVisible m_DelegateSetLabelVisible;
     ...

     public YourClass()
     {
           m_DelegateSetLabelVisible = new DelegateSetLabelVisible(this.DelegateSetLabelVisible);
     }


     private void DelegateSetLabelVisible(string strControlName, bool boolVisible)
     {
      Control ctrl = GetControlAtForm(this, strControlName);

      if (ctrl != null)
      {
            Object objCtrl = ctrl;
            if (objCtrl.GetType().ToString() == "System.Windows.Forms.Label")
            {
                  Label _label = (Label)objCtrl;
                  _label.Visible = boolVisible;
            }
      }
     }

    private Control GetControlAtForm(Control ctrl, string strControlName)
   {
      Control ListViewExCtrl = null;

      foreach (Control _ctrl in ctrl.Controls)
      {
            if (_ctrl.Name == strControlName)
            {
                  return(_ctrl);
            }
            if (_ctrl.HasChildren)
            {
                  Control ctrl_Temp = GetControlAtForm(_ctrl, strControlName);
                  if (ctrl_Temp != null)      { ListViewExCtrl = ctrl_Temp; break; }
            }
      }

                return (ListViewExCtrl);
     }

}




then use

BeginInvoke(m_DelegateSetLabelVisible, new object[] { "your_label", true });

to Invoke the LabelChange from the Worker Thread to the UI Thread
0
 
LVL 1

Author Comment

by:Kahhoe
ID: 12315557
Thank you for this very important concept... The clue is the BeginInvoke method!! This also helped me in solving the problem while I need to add a control in another thread. For example need to add a new row to a grid..
0
 
LVL 1

Expert Comment

by:CrazyIvan007
ID: 12317042
As i said you should use Delegates with Invoke/BeginInvoke if you access the UI Thread from other Threads. That means every Control you wan't to modify/change/add should be Invoked. Without that it can run, but it can throw Exceptions. So its safer to use it.
0

Featured Post

On Demand Webinar: Networking for the Cloud Era

Did you know SD-WANs can improve network connectivity? Check out this webinar to learn how an SD-WAN simplified, one-click tool can help you migrate and manage data in the cloud.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

In order to hide the "ugly" records selectors (triangles) in the rowheaders, here are some suggestions. Microsoft doesn't have a direct method/property to do it. You can only hide the rowheader column. First solution, the easy way The first sol…
This article introduced a TextBox that supports transparent background.   Introduction TextBox is the most widely used control component in GUI design. Most GUI controls do not support transparent background and more or less do not have the…
NetCrunch network monitor is a highly extensive platform for network monitoring and alert generation. In this video you'll see a live demo of NetCrunch with most notable features explained in a walk-through manner. You'll also get to know the philos…
Do you want to know how to make a graph with Microsoft Access? First, create a query with the data for the chart. Then make a blank form and add a chart control. This video also shows how to change what data is displayed on the graph as well as form…
Suggested Courses
Course of the Month13 days, 17 hours left to enroll

801 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question