C# Define base class by

Question

I am trying to find a way to derive a class from a generic base class. Say:

sealed public class Final : Base
{

}

public class Base

        

Answer 1

The result I'd like to get is to be able to do something like that:

 Anything[] anything;
 //Assign some Instances to anything 

 foreach(Final final in anything){
     //do something with final
 }

Your foreach loop suggests this: class Anything : Final { … }.

This obviously turns around the inheritance hierarchy as you planned and named it. (You cannot have cycles in your inheritance relationships).

---

public class Base<T> : T where T : Anything { …

Let me elaborate on this part for a bit. I'll reduce your example even further to just class Base<T> : T.

This is not possible, for good reason. Imagine this:

class Base<T> : T
{
    public override string Frobble()
    {
        Fiddle();
        return "*" + base.Frobble() + "*";
    }
}

class A
{
    public sealed string Frobble() { … }
}

class B
{
}

class C
{
    public virtual string Frobble() { … }
}

abstract class D
{
    public abstract void Fiddle();
    public virtual string Frobble() { … }
}

class E
{
    public void Fiddle() { … }
    public virtual string Frobble() { … }
}

You get all kinds of absurd situations if class Base<T> : T were allowed.

• Base<A> would be absurd because Frobble cannot be overridden in a derived class.
• Base<B> would be absurd because you cannot override a method that doesn't exist in the base class.
• Base<C> doesn't work because there is no Fiddle method to call.
• Base<D> would not work because you cannot call an abstract method.
• Only Base<E> would work.

How would the compiler ever know how to correctly compile Base<T> and analyse code that depends on it?

The point is that you cannot derive from a class that is not known at compile-time. T is a parameter, i.e. a variable, a placeholder. So class Base<T> : T is basically like saying, "Base<T> inherits from some (unknown) class". Class inheritance is a type relationship that requires both involved types to be known at compile-time. (Actually, that's not a super-precise statement because you can inherit from a generic type such as class SpecialList<T> : List<T>. But at the very least, the derived class has to know what members (methods, properties, etc.) are available in the base class.)

Answer 2

Is this what you want?

sealed public class Final : Base<int>{

}

public class Base<T> {

}

Answer 3

I don't know the context of this question, but I ran into same question with a project where I had to make it possible to extend the base class which is already derived by many others. Like:

abstract class Base {}
class FinalA : Base {}
class FinalB : Base {}

// Now create extended base class and expect final classes to be extended as well:
class BetterBase : Base {}

The solution was to create common ancestor and connect through properties:

abstract class Foundation {}
abstract class Base : Foundation 
{
    Foundation Final { get; }
}
class FinalA : Foundation {}
class FinalB : Foundation {}
class FinalC : Foundation
{
    Foundation Base { get; }
}

// Here's the desired extension:
class BetterBase : Base {}

Now BetterBase has connection to final class and if needed, the final classes could have connection with (Better)Base also, as shown in FinalC class.

Answer 4

You could only do this if Final would be a generic class as well, like so:

public sealed class Final<T> : Base<T>

Then you can put a type restraint on T as either a class, to allow only reference types as T, or an instance of Base<T>, to allow only types that derive from Base<T>:

public class Base<T> where T : Base<T>