Styjun

I'm using 2 forms in an application in which, clicking button in form1 opens form2. While opening form2, the Modal property of form2 is FALSE. But, once on clicking the button in form2, the value of this property is set to TRUE. I have used the below code in the application.

Form1

private void button1_Click(object sender, EventArgs e)

Form2 f2 = new Form2();
bool isModal = f2.Modal;
f2.ShowDialog();

Form2

private void button1_Click(object sender, EventArgs e)

bool isModal = this.Modal;

Kindly share your ideas on how this is updated?

Thanks,

Sindhu

The docs state:

Gets a value indicating whether this form is displayed modally.

The key word here is is.

Let's look at your code:

Form2 f2 = new Form2();
bool isModal = f2.Modal;
f2.ShowDialog();

The question you need to ask is "in that second line, is the form at that time shown modally?"

The answer is clearly No, since it is only shown modally on the third line. If you think logically, this makes perfect sense. The form doesn't know whether you are going to call Show or ShowDialog - so Modal can't tell you about the future - it can only tell you about the current state of affairs.

Thus, on the second line, Modal must (according to the docs) return false.

OK, so why is it true in here?

private void button1_Click(object sender, EventArgs e)

bool isModal = this.Modal;

Well, based on the docs we need to ask ourselves whether this is currently displayed modally when the button is clicked. Yes at that time it is, so it must (according to the docs) return true.

While it may appear to have something to do with the buttons, it does not. It has to do with ShowDialog(); method.

If you check before f2.ShowDialog();, the Modal will be false. Check after and it will be true.

This f2.ShowDialog(); is what sets the property to true.

You may use f2.Show() and it will stay false in this case for obvious reason.

As @CodingYoshi said, the method that is setting the Form.Modal to true is Form.ShowDialog(). This is why f2.Modal is false, because it is called before f2.ShowDialog().

The problem is checking the f2.Modal after calling f2.ShowDialog. The problem is that you cannot execute any further code lines in the code block after calling f2.ShowDialog(). Thus, there is no way to call f2.Modal.

The best way to see this difference to check Modal first in your constructor and then check it again in the Form.Load event. The constructor is called before ShowDialog is called, but Form.Load and Button.clicked are called after ShowDialog is called. Thus, you have different values for the Modal property.

Here the constructor of Form2:

Sub New ()
 ' This call is required by the designer.
 InitializeComponent()

 Console.WriteLine(Me.Modal) ' is always false
End Sub

And then Form.Loading event of Form2:

Private Sub Form2_Load(sender As Object, e As EventArgs) Handles MyBase.Load
 Console.WriteLine(Me.Modal) ' will return true if object is called via ShowDialog()
End Sub