CODEWITHSUNDEEP
c#winforms
MarisP
MarisP

Reputation: 987

Opening a new form via label causes Error to appear

I'm trying to make a new form appear when I click on a label. I'm using Windows Application Forms.

Here's the code:

private void label1_Click(object sender, EventArgs e)
{
   Form parpokeru = new Form();
   parpokeru.Show();
   parpokeru.ShowDialog();
}

When I click on the label, a error appears (Unhandled exception has occurred in your application...). Can anyone tell me how to fix it?

Upvotes: 0

Views: 105

Answers (2)

Steve
Steve

Reputation: 216293

When you call the Show method your form is shown on video like another window and the code returns immediately from the call. So your code continues and all the forms of your application are available for the user interaction. It is calledd a modeless dialog

On the contrary, ShowDialog is a blocking call. The code doesn't return from this call until something happen inside the called form that terminates the visualization of the form. As an example comes to ming a call to the methid Hide or a click on a button with its DialogResult property set to something different from DialogResult.None. At this point the code from ShowDialog returns and the normal processing continue. While the code is blocked inside the ShowDialog the application is blocked and the user cannot interact with other forms or menus or whatever is displayed on video. It is called modal dialog

Another difference is ShowDialog returns value (a DialogResult enum value) that can can be used to determine how the user closed the form (DialogResult.Cancel, DialogResult.OK), also ShowDialog does not call the Dispose method at closing time. This will allow to retrieve property from the Modal Dialog like user inputs for further processing.

I cannot imagine what happen in the internal processing of your form if, after a modeless call to Show you call immediately a ShowDialog on the same form instance. However, an exception is really the minimum to expect from this code.

Upvotes: 0

Grant Winney
Grant Winney

Reputation: 66449

Call .Show() or .ShowDialog(). Not both.

  • Show() will display your second form, while still allowing the user to access the first form.

  • ShowDialog() will display your second form as "modal". Execution of code in the first form stops while the second form is open (at least on the main thread.. for example, timers will continue to run), and the user will not be able to access your first form while the second is open.

Upvotes: 2

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