Question

I have the following base abstract class defined as:

public abstract class BaseObject<T> : IComparable, IComparable<T>, IEquatable<T> {}

I also have an interface defined as:

public interface ICode<T> where T : struct
{
    T Code { get; }
}

Now I want to derive a class that is inherited from BaseObject<T> and includes interface ICode<T>.

I tried to define it like that:

public class DerivedObject<T, U> : BaseObject<T>, ICode<U> where T : DerivedObject<T, U> where U : struct
{

    public DerivedObject(U code)
    {
        Code = code;
    }

    // From BaseObject
    protected override int InstanceCompareTo(T obj)
    {
        return Code.CompareTo(obj.Code);
    }

    // From BaseObject
    protected override bool InstanceEquals(T obj)
    {
        return Code.Equals(obj.Code);
    }

    // From ICode
    U _Code;

    public U Code 
    { 
        get { return _Code; }
        protected set { _Code = value; }
    }
}

The only error that comes from the compiler is for Code.CompareTo(obj.Code) with the message:

‘U’ does not contain a definition for ‘CompareTo’ and no extension method ‘CompareTo’ accepting a first argument of type ‘U’ could be found.

But U is a value type and should know CompareTo.

Have you any idea what I am doing wrong, or if I do all wrong?

My final aim is to derive classes such these:

    public class Account : DerivedObject<Account, int>
    public class ItemGroup : DerivedObject<ItemGroup, string>

Big Thanks In Advance!

Answer 1

It is not necessary for a struct type to have a CompareTo method. You have to enforce it using constraint:

public class DerivedObject<T, U> : BaseObject<T>, ICode<U>
    where T : DerivedObject<T, U>
    where U : struct, IComparable<U> {}

CompareTo method is part of IComparable<T> contract. And it has nothing to do with U being a value type. That is why, you have to add a generic constraint that states U is IComparable<U>.