form trouble

Question

Hello In opening form 1 I open form 2 in the constructor before I initialize componentss(). I created a public variable to set a int value in form2 so that if the correct value is returned form 1 continues on its way otherwise the project closes. I recieve this error when the retuned value is not set to 1 when i want the project to close. Is there another way that works? Is there a way to create a int value return for form 1? HelPPP! THanks. Make sure you have not released a resource before attempting to use it.  public Form 1() { Form2 frm2 = new Form2(); frm2.ShowDialog(); if (frm2.returned_value == 1) initializeComponents(); else{this.close();} }

j_sen21 · Answer

Perfect. Environment.Exit(1) did exactly what I wanted it to do. THanks

Alan · Answer

Well it's a bit drastic but you can avoid placing Application.Run() in a try block by replacing this.Close() with Environment.Exit(1). This will kill the process stone dead. It's not normally recommended for Windows forms applications but as Application.Exit() doesn't work it's probably all you can use here.

j_sen21 · Answer

THanks for the reply Alan I actually threw that in there but now I wont know when any other errors occur in form 1.

j_sen21 · Answer

Thanks Ryan for the response however the error still stands when canceling form1 from continuing. I guess the problem is more how to deal with the error.

j_sen21 · Answer

THanks for the reply Jan. I should have included that. I did that at first. But when you close down form 1 it leaves the program open when you check run programs. Unless there is a way of killing the whole app from form 1??? THanks

Alan · Answer

Although I prefer Jan's and Ryan's suggestions it is actually possible to do what you want by having a Returned_Value field in Form1 which is set from Form2. When Form2 is constructed you can pass a reference to Form1 as a parameter to enable the return value to be set. If the main form is closed in the constructor then the Application.Run() method in Main() which is responsible for displaying the main form will throw an exception because the main form instance has been disposed of. You can deal with this by calling Application.Run() within a try statement catching the exception and letting it fall through. This short (and heavily simplified) program will show you what I mean: using System; using System.Windows.Forms; class Test { static void Main() { try { Application.Run(new Form1()); } catch { } } } class Form1 : Form { public int Returned_value = 0; public Form1() { Form2 frm2 = new Form2(this); frm2.ShowDialog(); if (Returned_value == 1) this.Text = Form1; else this.Close(); } } class Form2 : Form { public Form2(Form1 frm1) { this.Text = Form2; // in practice you'd want to cache frm1 in a field // so it can be called from any method in Form2 frm1.Returned_value = 2; // change to '1' to load Form1 } }

Ryan Alford · Answer

why don't you use form2's dialog result attribute? then do... if (form2.showdialog() == dialogresult.OK) // depending on what the 1 means { // do code here }

Jan Montano · Answer

Hi Maybe it would be better if you set Form2 as your startup form instead. Then from Form2 depending if the condition is successful you'll call Form1. Cheers Jan