C#

public class Class1
	{
		public enum Protocol
		{
			Digital,
			Analog
		}
		public enum Model
		{
			Digital1,
			Digital2,
			Analog1,
			Analog2
		}

		public static Protocol m_selectedProtocol;
		public static Model m_ModelSelected;

		public static Protocol SelectedProtocol
		{
			get { return Class1.m_selectedProtocol; }
			set { Class1.m_selectedProtocol = value; }
		}
		public static Model ModelSelected
		{
			get { return Class1.m_ModelSelected ; }
			set { Class1.m_ModelSelected= value; }
		}

	}

private void button1_Click(object sender, EventArgs e)
{
    Class1.SelectedProtocol = Class1.Protocol.Analog;
}

if (rdAnalog.Checked == true)
 {
    Class1.SelectedProtocol = Protocol.Analog;


    if (Analog1.Checked == true)
    {
        Class1.ModelSelected = Model.Analog1;
    }
    else if (Analog2.Checked == true)
    {
        Class1.ModelSelected = Model.Analog2;
    }
    else
    {
        return;
    }
 }  /*This Curly bracket you have missed */
 else
 {
    return;
 }

Abydosgater wrote:

if you are using multiple forms, you needed to use Application.Run()?

Abydosgater wrote:

use form.Show()

Abydosgater wrote:

multi-threading

LoginForm loginForm = new LoginForm();<br />
                loginForm.ShowDialog();

<br />
public static void OpenMainForm(string accountname)<br />
{ Application.Run(new MainForm(accountname)); }

public class Form2 : Form {
	static void Main() {
		Application.EnableVisualStyles();
		Application.SetCompatibleTextRenderingDefault(false);
		new Form2("a").Show();
		new Form2("b").Show();
		Application.Run(new Form2("c"));
	}
	public Form2(string title) {
		Text=title;
	}
}

new Form2("b").Show();<br />
Application.Run(new Form2("c"));

Abydosgater wrote:

form.Show() only works if you have Application.Run on another form.

new Form2("b").Show();
new Form2("c").Show();
Application.Run();