OpenFileDialog doesn't show

Question

I have this simple code: ``` private void buttonOpen_Click(object sender, EventArgs e) { if (openFileDialog1.ShowDialog() == DialogResult.OK) { textBox2.Text = openFileDialog1.FileName; } } ``` When I run program form doesn't show and exit of debugging mode. In output view writes:The program '[4244] openfiledialog.vshost.exe: Managed (v4.0.30319)' has exited with code 1073741855 (0x4000001f). I have Visual Studio 2010 Professional. Edit:form1.designer.cs ``` private void InitializeComponent() { this.openFileDialog1 = new System.Windows.Forms.OpenFileDialog(); this.buttonOpen = new System.Windows.Forms.Button(); this.textBox1 = new System.Windows.Forms.TextBox(); this.textBox2 = new System.Windows.Forms.TextBox(); this.SuspendLayout(); // // openFileDialog1 // this.openFileDialog1.FileName = "openFileDialog1"; // // buttonOpen // this.buttonOpen.Location = new System.Drawing.Point(13, 48); this.buttonOpen.Name = "buttonOpen"; this.buttonOpen.Size = new System.Drawing.Size(75, 23); this…

Accepted Answer

As a general rule I initialize and use my OpenFileDialog's within the event that is calling it. I can't think of a circumstance in which I would want it to be a property of my window. The first thing I would do is delete it as a property and initialize it in your event.

```
private void buttonOpen_Click(object sender, EventArgs e)
{
    using (OpenFileDialog openFileDialog1 = new OpenFileDialog())
    {
        if (openFileDialog1.ShowDialog() == DialogResult.OK)
        {
            textBox2.Text = openFileDialog1.FileName;
        }
    }
}
```

You don't need to set the FileName property to anything because the dialog will do it for you.

The only thing I found on your error code was this (Program and debugger quit without indication of problem). Which in your current code I cannot find anything that would cause this. If you are accessing unmanaged code you may need to enable unmanaged code debugging.

OpenFileDialog doesn't show

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