在BackgroundWorker中调用ShowDialog

Question

I have a WinForms application in which my background worker is doing a sync task, adding new files, removing old ones etc.

In my background worker code I want to show a custom form to user telling him what will be deleted and what will be added if he continues, with YES/NO buttons to get his feedback.

I was wondering if it is ok to do something like this in background worker's doWork method? If not, how should I do it?

Please advise..

private void backgroundWorker1_DoWork(object sender, DoWorkEventArgs e)
{
   MyForm f = new MyForm();
   f.FilesToAddDelete(..);
   DialogResult result = f.ShowDialog();
   if(No...)
   return;
   else
   //keep working...
}

Answer 1

If you try this you will see for yourself that it will not work because the BackgroundWorker thread is not STA (it comes from the managed thread pool ).

The essence of the matter is that you cannot show user interface from a worker thread¹, so you must work around it. You should pass a reference to a UI element of your application (the main form would be a good choice) and then use Invoke to marshal a request for user interaction to your UI thread. A barebones example:

class MainForm
{

    // all other members here

    public bool AskForConfirmation()
    {
        var confirmationForm = new ConfirmationForm();
        return confirmationForm.ShowDialog() == DialogResult.Yes;
    }
}

And the background worker would do this:

// I assume that mainForm has been passed somehow to BackgroundWorker
var result = (bool)mainForm.Invoke(mainForm.AskForConfirmation);
if (result) { ... }

---

¹ Technically, you cannot show user interface from a thread that is not STA. If you create a worker thread yourself you can choose to make it STA anyway, but if it comes from the thread pool there is no such possibility.

Answer 2

I usually create a method to execute a delegate on the UI thread:

  private void DoOnUIThread(MethodInvoker d) {
     if (this.InvokeRequired) { this.Invoke(d); } else { d(); }
  }

With this, you can change your code to:

private void backgroundWorker1_DoWork(object sender, DoWorkEventArgs e)
{
   DialogResult result = DialogResult.No;
   DoOnUIThread(delegate() {
      MyForm f = new MyForm();
      f.FilesToAddDelete(..);
      result = f.ShowDialog();
   });

   if(No...)
   return;
   else
   //keep working...
}

Answer 3

IMO answers stating that you should launch a thread to handle this are misguided. What you need is to jump the window back to the main dispatcher thread.

In WPF

public ShellViewModel(
    [NotNull] IWindowManager windows, 
    [NotNull] IWindsorContainer container)
{
    if (windows == null) throw new ArgumentNullException("windows");
    if (container == null) throw new ArgumentNullException("container");
    _windows = windows;
    _container = container;
    UIDispatcher = Dispatcher.CurrentDispatcher; // not for WinForms
}

public Dispatcher UIDispatcher { get; private set; }

and then, when some event occurs on another thread (thread pool thread in this case):

public void Consume(ImageFound message)
{
    var model = _container.Resolve<ChoiceViewModel>();
    model.ForImage(message);
    UIDispatcher.BeginInvoke(new Action(() => _windows.ShowWindow(model)));
}

WinForms equivalent

Don't set UIDispatcher to anything, then you can do have:

public void Consume(ImageFound message)
{
    var model = _container.Resolve<ChoiceViewModel>();
    model.ForImage(message);
    this.Invoke( () => _windows.ShowWindow(model) );
}

DRYing it up for WPF:

Man, so much code...

public interface ThreadedViewModel
    : IConsumer
{
    /// <summary>
    /// Gets the UI-thread dispatcher
    /// </summary>
    Dispatcher UIDispatcher { get; }
}

public static class ThreadedViewModelEx
{
    public static void BeginInvoke([NotNull] this ThreadedViewModel viewModel, [NotNull] Action action)
    {
        if (viewModel == null) throw new ArgumentNullException("viewModel");
        if (action == null) throw new ArgumentNullException("action");
        if (viewModel.UIDispatcher.CheckAccess()) action();
        else viewModel.UIDispatcher.BeginInvoke(action);
    }
}

and in the view model:

    public void Consume(ImageFound message)
    {
        var model = _container.Resolve<ChoiceViewModel>();
        model.ForImage(message);
        this.BeginInvoke(() => _windows.ShowWindow(model));
    }

Hope it helps.

Answer 4

You should bring up the dialog before you run the backgroundworker. And in the progresschanged-event, you can update the dialog.