Singleton pattern for MDI child forms

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Introduction

When you make a MDI (Multi Document Interface) application, you usually need to have only one instance of each child-form to be open at any point of time. You can loop through all the open forms before showing a new form, and make sure that the form has not been opened before. In this article, I show you how you can use singleton pattern to accomplish the same result.

Using the code

The following steps will show you how to make a MDI child form using Singleton pattern.

1. Make MDI.cs, Form1.cs, and Form2.cs forms. (Set MDI.cs as MDI container.)
2. Add a menu bar and make two menu items: menuItem1, menuItem2.

If you open form1.cs in code view mode, you should see:

public Form1()
{
    //
    // Required for Windows Form Designer support
    //
    InitializeComponent();

    //
    // TODO: Add any constructor code after InitializeComponent call
    //
}

Since the constructor is public, you can make an instance of Form1 by using "new" keyword:

Form f = new Form1();

Every time you do this, it makes a new instance of Form1. Now, we should implement a logic that makes only one instance and returns that instance wherever Form1 is called.

To do so, make these changes:

private static Form1 aForm= null;
public static Form1  Instance()
{
    if(aForm==null)
    {
        aForm= new Form1();
    }
    return aForm;
}

private Form1()
{
    //
    // Required for Windows Form Designer support
    //
    InitializeComponent();

    //
    // TODO: Add any constructor code after InitializeComponent call
    //
}

Now, every time you need to access an instance of Form1, simply call its Instance() method, which is a public static member:

Form f = Form1.Instance();

First time you call this method, it checks the value of aForm variable, if it is null, it makes an instance of Form1 and assigns it to aForm, which is a static variable. After that, any time you call the Instance() method, it gets the object from the aForm instead of creating a new instance.

Now, let's map menuItem1 to Form1 and menuItem2 to Form2. Double click on menu items, then put the logic in:

private void menuItem1_Click(object sender, System.EventArgs e)
{
    Form f = Form1.Instance();
    f.MdiParent = this;
    f.Show();
    f.Activate();
}

private void menuItem2_Click(object sender, System.EventArgs e)
{
    Form f = new Form2();
    f.MdiParent = this;
    f.Show();
}

There is one more important thing that you need to do. Since aForm is a static variable, if you close the Form1, the value of aForm won't get reset automatically, so you need to do a clean up. To do this, add "aForm = null;" to the dispose method of Form1:

protected override void Dispose( bool disposing )
{
    if( disposing )
    {
        if(components != null)
        {
            components.Dispose();
        }
    }
    base.Dispose( disposing );
    aForm = null;
}

Now, you can run the application. Every time you click on menuItem2, it opens a new instance of Form2, but you can only open one instance of Form1.