C#

public Settings setting = new Settings();

settings = new Settings();
settings.ShowDialog();
...

samskiter wrote:

I have declared my form as a class member variable ... because i want to only be able to show it once

try
            {
                this.setting.Activate();
            }
            catch
            {
                this.setting = new Settings();
                this.setting.Show();
            }

public sealed class SingletonForm : Form
{
    SingletonForm() { }
    public static SingletonForm Instance
    {
        get { return Nested.instance; }
    }
    private class Nested
    {
        static Nested() { }
        internal static readonly SingletonForm instance = new SingletonForm();
    }
}

public partial class Settings : Form<br />
    {<br />
        public Settings()<br />
        {<br />
            InitializeComponent();<br />
        }<br />
        public static Settings Instance<br />
        {<br />
            get { return Nested.instance; }<br />
        }<br />
        private class Nested<br />
        {<br />
            static Nested() { }<br />
            internal static readonly Settings instance = new Settings();<br />
        }<br />
    }

public sealed partial class Settings : Form
{
    private Settings()
    {
        InitializeComponent();
    }
    public static Settings Instance
    {
        get { return Nested.instance; }
    }
    private class Nested
    {
        static Nested() { }
        internal static readonly Settings instance = new Settings();
    }
}

public void COMMenuItem_Click(object sender, EventArgs e)
        {
            Settings.Instance.ShowDialog();
        }

samskiter wrote:

is allowed to be called even when the form is already there

public void OldMain()
        {
            if (SerPort.OpenPort())
            {
                icn = new NotifyIcon();
                ContextMenuStrip menu = new ContextMenuStrip();
                icn.ContextMenuStrip = menu;

                ToolStripMenuItem COMMenuItem = new ToolStripMenuItem("COM Port", null, new EventHandler(this.COMMenuItem_Click));
                ToolStripMenuItem ExitMenuItem = new ToolStripMenuItem("Exit", null, new EventHandler(ExitMenuItem_Click));
                ToolStripSeparator separator = new ToolStripSeparator();
                menu.Items.AddRange(new System.Windows.Forms.ToolStripItem[] { COMMenuItem, separator, ExitMenuItem });
                icn.Visible = true;
                icn.Icon = Properties.Resources.ehshell_IDI_APP1;
                icn.ShowBalloonTip(3000, "Connected", "Duke Box is Connected", ToolTipIcon.Info);
                //could add code for "not show this tooltip message in future"
                MediaCenterController MCC = new MediaCenterController();
                MCC.MediaStateChange += new MediaCenterController.MediaCenterHandler(SerPort.MediaCenter_Running);
                MCC.MediaStateChange += new MediaCenterController.MediaCenterHandler(MediaCenter_Running);
                Thread trd = new Thread(new ThreadStart(MCC.MediaCenterRunning));
                trd.IsBackground = true;
                trd.Start();
                
                Application.Run();
            }
            else
            {
                //could create an event for new ports detection
            }
        }

samskiter wrote:

yes, its a notifyicon

private void COMMenuItem_Click(object sender, EventArgs e)
{
    COMMenuItem.Enabled = false;
    // other stuff here if needed
    Settings.ShowDialog();
    // more stuff here if needed
    COMMenuItem.Enabled = true;
}