Open form based on int

Question

I'm just starting Project Euler and I have already ran into my first issue. I want to have a separate form for every Euler Problem but I cannot figure out how to open the forms in a good way. What I want to do is use the variable problemNumber to open a form. For example if I have problemNumber set to 54 and I press a button it should open form54.

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Windows.Forms;

namespace WindowsFormsApplication1
    {
    public partial class Form1 : Form
    {
        public Form1()
        {
            InitializeComponent();
        }

        private void button1_Click(object sender, EventArgs e)
        {
             int problemNumber = int.Parse(numericUpDown1.Text);


        }
    }
}

I do know how to open a specific form, for example form2.

form2 f2 = new form2();
f2.Show();

This would open up form2, but as mentioned above I want to open up form+problemNumber.

Thank you!

Answer 1

One possibility would be to create a static class, that has one static method with a switch case and returns a Form object. In the method you create new form object regarding the specified number:

public static class FormStarter
{
    public static Form OpenForm(int formNumber)
    {
        Form form = null;
        switch (formNumber)
        {
            case 1: form = new Form1(); break;
            case 2: form = new Form2(); break;
            //case 3: form = new Form3(); break;
            // ...
            default: throw new ArgumentException();
        }

        return form;
    }
}

You can then use the class like this:

var f = FormStarter.OpenForm(2);
f.ShowDialog(); // Form2 is started!

Once you have a new form, you need to add the creation of the instance in only one method - the OpenForm.

Another possible solution without a static class would be a Dictionary object holding the number and the form instance like this:

Dictionary<int, Form> forms = new Dictionary<int, Form>();
forms.Add(1, new Form1());
forms.Add(2, new Form2());
forms[2].ShowDialog(); // Form2 is started!

If you want to reference just the type of the form in the dictionary and create an instance only when needed, then change the value of the dictionary to be of type Type and put the corresponding form in it:

Dictionary<int, Type> forms = new Dictionary<int, Type>();
forms.Add(1, typeof(Form1));
forms.Add(2, typeof(Form2));
((Form)Activator.CreateInstance(forms[2])).ShowDialog(); // Form2 is started!