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Home/ Questions/Q 8794045
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T23:12:00+00:00 2026-06-13T23:12:00+00:00

Probably I don’t understand correctly this topic. Here is an issue… C# Windows Application

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Probably I don’t understand correctly this topic. Here is an issue…

C# Windows Application (.NET 2.0). The MainForm has a button “Query”. When the user pushes it, the following should happen:

private void btnQuery_Click(object sender, EventArgs e)
{
    querier = new Querier();
    OutputForm outputForm = new OutputForm();
    querier.ProcessAll(outputForm.OutputReceived);
    outputForm.ShowDialog();
}

Querier is the worker. It creates a background thread and runs it to do stuff. The OutputForm is a simple form with txtOutput multiline text-box that should display the output of the working thread.

To allow the working thread send its output, querier.ProcessAll() method receives a callback handler. This is its implementation:

public void OutputReceived(string message)
{
    if (this.InvokeRequired)
        this.Invoke((MethodInvoker)delegate() { this.OutputReceived(message); });
    else if (!string.IsNullOrEmpty(message))
        txtOutput.AppendText(message + Environment.NewLine);
}

So basically the working thread runs and sends output using the OutputReceived() method, which uses Invoke(), because the working thread can’t access the txtOutput field directly.

Note that outputForm.ShowDialog() is called AFTER querier.ProcessAll(). That’s because ShowDialog() is blocking.

But here is the problem. If the working thread sends any output BEFORE the dialog is actually shown, I get the exception about cross-thread operation! When I debug it, I see that for some reason this.InvokeRequired() in the OutputReceived() method returns "false"! That’s why the working thread tries to access txtOutput directly and crashes.

The problem is clearly about the race condition between the thread and ShowDialog(). If I add Thread.Sleep() in the beginning of the working thread, the dialog shows up and then everything works fine.

Can you explain such behavior?

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  1. Editorial Team
    Editorial Team
    2026-06-13T23:12:01+00:00Added an answer on June 13, 2026 at 11:12 pm

    Sorry, I found the answer!

    There is one special case when InvokeRequired() will return "false". It’s when the control’s handle hasn’t been created yet. In this case it is forbidden to call Invoke() – so the InvokeRequired() tries to protect you, sort of.

    Now I call CreateHandle() method in the CTOR of OutputForm. In this way the handle is created even before the dialog is shown, so InvokeRequired() works as expected.

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