Closing Forms in Windows Form App?

Open second form in dialog mode:

this.Hide(); B.ShowDialog(); this.Show(); 

lan, the problem with your code is that you don't return to the old formA when you press btnReopenA in formB, instead you open a new formA.

an elegant way to avoid that will be to register to the FormClosing event

private void btnOpenformB_Click(System.Object sender, System.EventArgs e) { FormB B = new FormB(); B.FormClosing += b_FormClosing; this.Hide(); B.Show(); } void b_FormClosing(object sender, FormClosingEventArgs e) { Show(); } 

or, if you don't want to deal with events you can do this:

public partial class FormB : Form { private Form _frm; public FormB(Form frm) { _frm = frm; InitializeComponent(); } private void btnReopenA_Click(System.Object sender, System.EventArgs e) { if(_frm!=null) _frm.Show(); this.Close(); } } 

and when creating formB:

private void btnOpenformB_Click(System.Object sender, System.EventArgs e) { FormB B = new FormB(this); this.Hide(); B.Show(); } 

Use Application.OpenForms[] collection: Application.OpenForms["FormA"].Show()

My problem is when I click the button on FormB to reopen FormA, and when I click the exit button on FormA, it doesn't stop debugging.

This is because you opened another instance of FormA form your FromB

What should I do?

You need to get a reference of FormA in FormB and Show that instead.

How do i do it?

 public class FormA { private void btnOpenformB_Click(System.Object sender, System.EventArgs e) { FormB B = new FormB(); B.Closed+=OnFromBClosed; //Add this to handle FromB Closed event this.Hide(); B.Show(); } private void btnExit_Click(System.Object sender, System.EventArgs e) { this.Close(); } //Show FormA again when FromB is closed protected void OnFromBClosed(object sender, EventArgs e) { this.Show(); } } public class FormB { private void btnReopenA_Click(System.Object sender, System.EventArgs e) { // FormA A = new FormA(); remove this. this.Close(); // A.Show(); and remove this } }