The method or operation is not implemented

Question

There are two forms. Form2 is derived from Form1.

But I have an issue with Form2 in design mode as shown on the screenshot below.

If I will comment this this._presenter.Retrive(); it will work fine. Whats going on and how to solve the problem?

UPD: If I will remove the throw new NotImplementedException(); and will insert, for example, MessageBox.Show("Test");, every time I will open Form2 the MessageBox will appears as though I run the application.

Form2

namespace InheritanceDemo
{
    public partial class Form2 : Form1
    {
        public Form2()
        {
            InitializeComponent();
        }
    }
}

Form1

namespace InheritanceDemo
{
    public partial class Form1 : Form
    {
        protected IPresenter _presenter;

        public Form1()
        {
            InitializeComponent();
            _presenter = new Presenters();
        }

        private void Form1_Load(object sender, EventArgs e)
        {
            this._presenter.Retrive();
        }
    }

    public class Presenters : IPresenter
    {
        public void Retrive()
        {
            throw new NotImplementedException();
        }
    }

    public interface IPresenter
    {
        void Retrive();
    }
}

Instead of `throw new NotImplementedException();` do something else.... :) — Fᴀʀʜᴀɴ Aɴᴀᴍ, Jan 02 '16 at 15:46
@FᴀʀʜᴀɴAɴᴀᴍ If I will remove the `throw new NotImplementedException();` and will insert, for example, `MessageBox.Show("Test");`, every time I will open Form2 the `MessageBox` will appears as though I run the application. — Mikhail Danshin, Jan 02 '16 at 16:00
By *something else* I meant some logic your interface is supposed to implement. Not just anything like MessageBox. — Fᴀʀʜᴀɴ Aɴᴀᴍ, Jan 02 '16 at 16:01
@FᴀʀʜᴀɴAɴᴀᴍ I understood. But why is there such a behavior? — Mikhail Danshin, Jan 02 '16 at 16:06
you will find [Can't view designer when coding a form in C#](http://stackoverflow.com/a/32299687/3110834) and [Show controls added programatically in WinForms app in Design view](http://stackoverflow.com/a/33535154/3110834) posts helpful and interesting! — Reza Aghaei, Jan 02 '16 at 16:10

Reza Aghaei · Accepted Answer · 2019-10-03T05:03:23.753

18

The main cause of error is what Fᴀʀʜᴀɴ and Yuval said:

throw new NotImplementedException();

But there is another important thing that you should pay attention to.

OP: If I will remove the throw new NotImplementedException(); and will insert, for example, MessageBox.Show("Test");, every time I will open Form2 the MessageBox will appears as though I run the application

If you notice, you will not receive this error in designer of Form1. But because your Form2 inherits from Form1 you receive this error.

It's because, when you open a form in designer, the designer makes an instance of base class of your form to show your form. It means instead of creating an instance of Form2 it creates an instance of Form1, runs Form1 constructor and hosts it in the design surface, and then deserializes the codes in InitializeComponent of Form2 and puts components on the design surface.

This is why you receive the error when you see your Form2 in designer, but you didn't receive any error while opening the Form1 in designer.

To solve the problem:

You can remove the implementation, and let the implementation be empty.
Also you can prevent the error by prevent running the code in Form_Load fd you are at design mode using DesignMode property, in Form1_Load:

if (DesignMode) return;

You probably will find these answers helpful and interesting:

edited Oct 03 '19 at 05:03

answered Jan 02 '16 at 16:02

Reza Aghaei

120,393
18
203
398

@FᴀʀʜᴀɴAɴᴀᴍ Sorry about misspelled name :) – Reza Aghaei Jan 02 '16 at 16:04
But what if my `DoSometning()` method is not working correctly right now. – Mikhail Danshin Jan 02 '16 at 16:12
What do you expect as result? – Reza Aghaei Jan 02 '16 at 16:13
I want to finish the application framework, then go and do the design, and then write the implementation. – Mikhail Danshin Jan 02 '16 at 16:16
You can remove the implementation, and let the implementation be empty. Also you can prevent the error by prevent running the code in `Form_Load` id you are at design mode. – Reza Aghaei Jan 02 '16 at 16:18
What do you mean? How can I prevent running the code at design mode? – Mikhail Danshin Jan 02 '16 at 16:21
What exactly is `DesignMode` in your example? – Mikhail Danshin Jan 02 '16 at 16:57
[`DesignMode`](https://msdn.microsoft.com/en-us/library/system.web.ui.control.designmode(v=vs.110).aspx) Gets a value indicating whether a control is being used on a design surface. – Reza Aghaei Jan 02 '16 at 17:04
@MikhailDanshin It seems the post answers your question and it would be great if you accept the post :) – Reza Aghaei Jan 09 '16 at 21:49

score 5 · Answer 2 · answered Jan 02 '16 at 15:54

Whats going on and how to solve the problem?

This is fairly trivial. If you would of debugged your code, you'd see that you're throwing a NotImplementedException in your method call, that is why commenting it out works:

public void Retrive()
{
    throw new NotImplementedException();
}

Instead of throwing, perhaps you want to implement the actual method logic.

score 1 · Answer 3 · answered Nov 09 '16 at 08:01

1

Commenting out the part throw new NotImplementedException(); worked fine for me Now, finally method will be alike below:

public void Retrive()
{
    //throw new NotImplementedException();
}

answered Nov 09 '16 at 08:01

Anjan Kant

4,090
41
39

The method or operation is not implemented

3 Answers3