How to continue executing code after calling ShowDialog()

Question

the Form.ShowDialog() method causes the code to be halted until the newly called form is closed. I need the code to continue running after the ShowDialog() method is called.

Answer 1

I suppose next solution for async ShowDialog:

public bool DialogResultAsync
{
    get;
    private set;
}

public async Task ShowDialogAsync()
{
    var cts = new CancellationTokenSource();
    // Attach token cancellation on form closing.
    Closed += (object sender, EventArgs e) =>
    {
        cts.Cancel();
    };
    Show(); // Show message without GUI freezing.
    try
    {
        // await for user button click.
        await Task.Delay(Timeout.Infinite, cts.Token);
    }
    catch (TaskCanceledException)
    { } 
}

public void ButtonOkClick()
{
    DialogResultAsync = true;
    Close();
}

public void ButtonCancelClick()
{
    DialogResultAsync = false;
    Close();
}

And in main form you must use this code:

public async void ShowDialogAsyncSample()
{
    var msg = new Message();
    if (await msg.ShowDialogAsync())
    {
        // Now you can use DialogResultAsync as you need.
        System.Diagnostics.Debug.Write(msg.DialogResultAsync);
    }
}