C# Modal dialog box (ShowDialog or MessageBox.Show) in async method not works as expected

Question

I have a topmost WinForm with a simple button that executes some commands asynchronously:

        private async void button1_Click(object sender, EventArgs e)
    {
        await System.Threading.Tasks.Task.Run(() =>
        {
            System.Threading.Thread.Sleep(2000);

            //Problem1: not works as "modal" dialog box (Main form remains active!) 
            new Form().ShowDialog();

            //Problem2: not shown as "modal" messagebox (Main form remains active!)
            MessageBox.Show("Test");
        });
    }

Inside the async function are the Messagebox.Show() and ShowDialog() methods, BUT:

Problem 1(solved): The new form does not open as modal dialog box (the main form is still active and accessible!)

Problem 2(solved): The MessageBox.Show() method doesn't behave as modal dialog box (the main form is still active and accessible.).

I need async-await to prevent the main UI from freezing, but I also want messageboxes (and sub-forms) inside to be displayed as modal dialog box. How can i show modal dialog boxes (on the main topmost Form) via async method?

Thanks

Answer 1

Solution for problem-1: ShowDialogAsync extension method solves the problem.

Solution for problem-2:

        private async void button1_Click(object sender, EventArgs e)
    {

        var handle = this.Handle;
        await System.Threading.Tasks.Task.Run(() =>
        {
            System.Threading.Thread.Sleep(2000);

            //Solution for Problem2:

            NativeWindow win32Parent = new NativeWindow();
            win32Parent.AssignHandle(handle);
            //Works as expected (Topmost and Modal):
            MessageBox.Show(win32Parent, "Test");
        });
    }

Related topic

Answer 2

Task.Run() runs work in background thread. In common you shouldn't show windows from background threads, but you can. To solve your problem you need to use UI thread Dispatcher.

Application.Current.Dispatcher.Invoke(() => MessageBox.Show("Test"));

Or

Application.Current.Dispatcher.Invoke(() => new Form().ShowDialog());