Question

This may seem like a silly question, but I just want to make sure I have it right. My main form is not visible most of the time. To open it I have a NotifyIcon. One of the menu options is Exit Application. I also have a few static global variables that need to be disposed of before the application closes. so in program.cs i have this.

    [STAThread]
    static void Main()
    {
        InitializeApplication();
        InitializeMainForm();
        Application.Run(main);
    }
    private static void InitializeApplication()
    {
        Application.EnableVisualStyles();
        Application.SetCompatibleTextRenderingDefault(false);
        Application.ApplicationExit += Application_ApplicationExit;
    }
    private static void InitializeMainForm()
    {
        main = new AssignButtonForm();
        main.FormClosing += main_FormClosing;
        Globals.StartNotify();
    }

    static void main_FormClosing(object sender, FormClosingEventArgs e)
    {
        var dlg = MessageBox.Show("Turn off Application?", "Exit?", MessageBoxButtons.OKCancel, MessageBoxIcon.Question, MessageBoxDefaultButton.Button1);
        if (dlg == DialogResult.OK)
        {
            Globals.notifyIcon1.Dispose();
            Application.Exit();
        }
        else
        {
            e.Cancel = true;
        }
    }

So what i’m hoping is that this is the correct way to call that.

    private void exitToolStripMenuItem_Click(object sender, EventArgs e)
    {
        this.Hide();
        Application.OpenForms[0].Close();
    }

is that correct? or is there a better way?

EDIT

ok so one of my classes that has its own Dispose method

public class KeymonNotifyIcon : IDisposable
{
    public KeymonNotifyIcon()
    {
        InitializeComponent();
        keymonMenuStrip.SetupKeysSelected += OnSetupKeysSelected;
    }
    ~KeymonNotifyIcon()
    {
        Dispose();
    }
    public void Dispose()
    {
        if (notifyIcon1 != null)
            notifyIcon1.Dispose();

        if (keymonMenuStrip != null)
            keymonMenuStrip.Dispose();
    }
}

The global class

public static class Globals
{
    public static TraceSource trace = new TraceSource("Keymon");
    public static KeymonNotifyIcon notifyIcon1;
    public static void StartNotify()
    {
        notifyIcon1 = new KeymonNotifyIcon();
    }
}

Answer 1

If your program just shuts down, objects that have not been garbage collected will have their finalizers run. It is the recommended pattern that objects implementing IDisposable have a finalizer to ensure that IDisposable is run. To the best of my knowledge, .NET BCL classes always follow that pattern. However, your own or third-party/open source components might not follow that pattern.

Note: The linked pattern on MSDN does not call GC.SuppressFinalize. Check out how to use it to reduce the GC overhead.

Implementing IDisposable alone isn’t sufficient to ensure that the object is properly disposed.

If you have statically referenced objects that implement IDisposable, it is a more solid solution to explicitly call IDisposable.Dispose() on them from an application shut down event.

EDIT

Your Dispose implementation will cause Dispose() to be called twice on the owned objects because:

if (notifyIcon1 != null)
        notifyIcon1.Dispose();

does not set notifyIcon1 to null, and the finalizer unconditionally calls Dispose() again.

Additionally, you will always cause the finalizer to run (which makes GC more expensive for that class) because you do not call GC.SuppressFinalize() in Dispose().