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

Edit: It's proberly not a good idea to change the functionality of ShowDialog(), people expect it to return a DialogResult and show the form, I suggest something like my suggestion below. Thus, still allowing ShowDialog() to be used the normal manner.

You could create a static method on your MyForm, something like DoShowGetResult()

which would look something like this

public static MyResultsForm DoShowGetResult()
{
   var f = new MyForm();
   if(f.ShowDialog() == DialogResult.OK)
   {
      return f.Result;   // public property on your form of the result selected
   }
   return null;
}

then you can use this

MyFormsResult result = MyForm.DoShowGetResult();