Is it possible to overload the ShowDialog method for forms and return a different result?

Question

EDIT: This method actually works great and I asked it then found the solution later. I added the correct call in the overloaded ShowDialog() method (it\'s not exacly an over

Answer 1

No, it's not possible. The usual workaround is to expose your real result as a property on the Form:

public MyFormResults MyResult
{
    get;
}

and you would then set this:

private void btn1_click(object sender, EventArgs e)
{
    MyResult = MyFormsResults.Result1;
    this.DialogResult = DialogResult.OK; //Do I need this for the original ShowDialog() call?
    this.Close(); //Should I close the dialog here or in my new ShowDialog() function?
}

and the calling code usually looks like this:

if (form.ShowDialog == DialogResult.OK)
{
    //do something with form.MyResult
}