The method or operation is not implemented
Solution 1
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 usingDesignMode
property, inForm1_Load
:if (DesignMode) return;
You probably will find these answers helpful and interesting:
- Can't view designer when coding a form in C#
- Show controls added programatically in WinForms app in Design view
Solution 2
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.
Solution 3
Commenting out the part throw new NotImplementedException(); worked fine for me Now, finally method will be alike below:
public void Retrive()
{
//throw new NotImplementedException();
}
Mikhail Danshin
MCP, MVP, Системный администратор, Программист на C#. Веду блог http://www.danshin.ms.
Updated on December 22, 2021Comments
-
Mikhail Danshin over 2 years
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(); } }
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Reza Aghaei over 8 years@FᴀʀʜᴀɴAɴᴀᴍ Sorry about misspelled name :)
-
Mikhail Danshin over 8 yearsBut what if my
DoSometning()
method is not working correctly right now. -
Reza Aghaei over 8 yearsWhat do you expect as result?
-
Mikhail Danshin over 8 yearsI want to finish the application framework, then go and do the design, and then write the implementation.
-
Reza Aghaei over 8 yearsYou 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. -
Mikhail Danshin over 8 yearsWhat do you mean? How can I prevent running the code at design mode?
-
Mikhail Danshin over 8 yearsWhat exactly is
DesignMode
in your example? -
Reza Aghaei over 8 years
DesignMode
Gets a value indicating whether a control is being used on a design surface. -
Reza Aghaei over 8 years@MikhailDanshin It seems the post answers your question and it would be great if you accept the post :)