Error when Implementing Interface: class doesn't implement interface members

Question

I am trying to implements the IUpdatable.

Error 1 'WebRole1.InfoManager' does not implement interface member 'System.Data.Services.IUpdatable.ClearChanges()' All the errors i get are saying that i am not implementing all interface members, but i implements some not all of course. I didn't put the hole code I hope that you can understand.

  using System;
  using System.Collections.Generic;
  using System.Linq;
  using System.Web;
  using System.Data.Services;
  using Microsoft.WindowsAzure;
  using Microsoft.WindowsAzure.ServiceRuntime;
  using Microsoft.WindowsAzure.StorageClient;

  namespace WebRole1
  {
     public class InfoManager : IUpdatable
     {
      private TableServiceContext context;

    // To Generate DataConnectionString and svcClient
    private TableServiceContext GetContext()
    {
    //Implemented code
    }

    public CommentManager()
    {
        context = GetContext();
    }


    // To get my table infos
    public IQueryable<Info> Infos
    {
        get
        {
            return context.CreateQuery<Info>("Infos").AsTableServiceQuery();
        }
    }
   // Creating the resource and cheking the compatibility of the type and do an add Object 

    public Object CreateResource(string containerName, string fullTypeName)
    {
        //Implemented Code
    }

    // Return the instance of the resource represented by the object 
    public Object ResolveResource(Object resource)
    {
        return resource;
    }

    public void SaveChanges()
    {
        context.SaveChangesWithRetries();
    }

    public void setValue(Object targetResource, string propertyName, Object propertyValue)
    {
    //Implemented Code
    }

}

}

Grant Thomas · Accepted Answer · 2011-03-18T15:23:30.570

8

It's an interface, therefore you MUST implement all members - whether you want to, or not.

This error will not go away until you implement the interface in its entirety. You can do what you want within the scope of the method you're implementing, say even raising a NotImplementedException for instance - but that is your implementation so the compiler is happy.

I won't condone ignorance (you should still learn the hows and whys) but I will offer a tip which may aid in your learning and, if nothing else, future productivity:

From within Visual Studio, when you have the class code file open which is to implement the interface then you can have VS spit out the code for you...

class MyClass : IMyInterface // <- hover mouse and click the drop down that appears

From the drop down you should see the option Implement Interface 'IMyInterface', click it and voilà! It will automatically generate the skeletal method bodies for you.

edited Mar 18 '11 at 15:23

answered Mar 18 '11 at 14:42

Grant Thomas

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ok so if i need only some methods i can raise the NotImplementedException, that is faster, thanks. – 404Dreamer_ML Mar 18 '11 at 14:49
While that is certainly the correct approach, it is not exactly a good habit to get into. As a coworker, I would always question why you're implementing an interface without all its methods. You are essentially lying, unless the interface dictates that such implementations are permissible. – jdmichal Mar 18 '11 at 14:53
If you want fast then see my update for auto-generating. If you want to do it by hand then defining the method body and omitting any exception raising would be faster than _adding_ a `throw`. – Grant Thomas Mar 18 '11 at 14:55

score 1 · Answer 2 · answered Mar 18 '11 at 14:42

1

I don't understand what the question is. If you implement an interface, you must implement all methods in that interface. Otherwise the compiler will give an error.

answered Mar 18 '11 at 14:42

jdmichal

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@404Dreamer_ML Yes. It will give errors unless you implement _all_ the methods. That is what implementing an interface means, that you implement all its methods. – jdmichal Mar 18 '11 at 14:54

score 1 · Answer 3 · answered Mar 18 '11 at 14:43

1

I think the error is clear: you are not implementing all interface members, which is required of course.

answered Mar 18 '11 at 14:43

Gerrie Schenck

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Error when Implementing Interface: class doesn't implement interface members

3 Answers3