Question

I have a SafeInvoke Control extension method similar to the one Greg D discusses here (minus the IsHandleCreated check).

I am calling it from a System.Windows.Forms.Form as follows:

public void Show(string text) {
    label.SafeInvoke(()=>label.Text = text);
    this.Show();
    this.Refresh();
}

Sometimes (this call can come from a variety of threads) this results in the following error:

System.InvalidOperationException occurred

Message= “Invoke or BeginInvoke cannot be called on a control until the window handle has been created.”

Source= “System.Windows.Forms”

StackTrace:
at System.Windows.Forms.Control.MarshaledInvoke(Control caller, Delegate method, Object[] args, Boolean synchronous)
at System.Windows.Forms.Control.Invoke(Delegate method, Object[] args)
at System.Windows.Forms.Control.Invoke(Delegate method)
at DriverInterface2.UI.WinForms.Dialogs.FormExtensions.SafeInvoke[T](T control, Action`1 action) 
in C:\code\DriverInterface2\DriverInterface2.UI.WinForms\Dialogs\FormExtensions.cs:line 16

What is going on and how do I fix it? I know as much as it is not a problem of form creation, since sometimes it will work once and fail the next time so what could the problem be?

PS. I really really am awful at WinForms, does anyone know a good series of articles that explains the whole model and how to work with it?

Answer 1

It’s possible that you’re creating your controls on the wrong thread. Consider the following documentation from MSDN:

This means that InvokeRequired can
return false if Invoke is not required
(the call occurs on the same thread),
or if the control was created on a
different thread but the control’s
handle has not yet been created.

In the case where the control’s handle
has not yet been created, you should
not simply call properties, methods,
or events on the control. This might
cause the control’s handle to be
created on the background thread,
isolating the control on a thread
without a message pump and making the
application unstable.

You can protect against this case by
also checking the value of
IsHandleCreated when InvokeRequired
returns false on a background thread.
If the control handle has not yet been
created, you must wait until it has
been created before calling Invoke or
BeginInvoke. Typically, this happens
only if a background thread is created
in the constructor of the primary form
for the application (as in
Application.Run(new MainForm()),
before the form has been shown or
Application.Run has been called.

Let’s see what this means for you. (This would be easier to reason about if we saw your implementation of SafeInvoke also)

Assuming your implementation is identical to the referenced one with the exception of the check against IsHandleCreated, let’s follow the logic:

public static void SafeInvoke(this Control uiElement, Action updater, bool forceSynchronous)
{
    if (uiElement == null)
    {
        throw new ArgumentNullException("uiElement");
    }

    if (uiElement.InvokeRequired)
    {
        if (forceSynchronous)
        {
            uiElement.Invoke((Action)delegate { SafeInvoke(uiElement, updater, forceSynchronous); });
        }
        else
        {
            uiElement.BeginInvoke((Action)delegate { SafeInvoke(uiElement, updater, forceSynchronous); });
        }
    }
    else
    {    
        if (uiElement.IsDisposed)
        {
            throw new ObjectDisposedException("Control is already disposed.");
        }

        updater();
    }
}

Consider the case where we’re calling SafeInvoke from the non-gui thread for a control whose handle has not been created.

uiElement is not null, so we check uiElement.InvokeRequired. Per the MSDN docs (bolded) InvokeRequired will return false because, even though it was created on a different thread, the handle hasn’t been created! This sends us to the else condition where we check IsDisposed or immediately proceed to call the submitted action… from the background thread!

At this point, all bets are off re: that control because its handle has been created on a thread that doesn’t have a message pump for it, as mentioned in the second paragraph. Perhaps this is the case you’re encountering?