Question

It seems that if you add an entry to a ContextMenu while it’s open, it’ll turn into a blank, tiny little square.

Like this:

To reproduce, just create a new WinForms application and replace the Form1 class with this:

public partial class Form1 : Form
{
    ContextMenu _menu = new ContextMenu();

    public Form1()
    {
        InitializeComponent();

        ContextMenu = _menu;

        _menu.MenuItems.Add(new MenuItem() { Text = "Test" });

        Timer a = new Timer() { Interval = 3000 };
        a.Tick += (sender, e) =>
        {
            _menu.MenuItems.Add(new MenuItem() { Text = "Woah!" });
        };
        a.Start();
    }
}

Then simply launch, right click and wait.

Is it possible to work around this without resorting to using something like ContextMenuStrip?

Answer 1

Use Show method right after addition to workaround the issue:

Timer a = new Timer() { Interval = 3000 };
a.Tick += (sender, e) =>
{
    _menu.MenuItems.Add(new MenuItem() { Text = "Woah!" });
    _menu.Show(this, Point.Empty);
};

You may also need to track the popup state to avoid unexpected showing. Here is the full source to achieve that:

public partial class Form1 : Form
{
    ContextMenu _menu = new ContextMenu();

    public Form1()
    {
        InitializeComponent();

        ContextMenu = _menu;

        _menu.Popup += new EventHandler(_menu_Popup);
        _menu.Collapse += new EventHandler(_menu_Collapse);

        _menu.MenuItems.Add(new MenuItem() { Text = "Test" });

        Timer a = new Timer() { Interval = 3000 };
        a.Tick += (sender, e) =>
        {
            _menu.MenuItems.Add(new MenuItem() { Text = "Woah!" });
            if (_menuPoppedUp)
                _menu.Show(this, Point.Empty);
        };
        a.Start();
    }

    bool _menuPoppedUp;

    void _menu_Collapse(object sender, EventArgs e)
    {
        _menuPoppedUp = false;
    }

    void _menu_Popup(object sender, EventArgs e)
    {
        _menuPoppedUp = true;
    }
}