使用不同类型参数实现接口的两个类

Question

class SingleTon{
   Data<A> a;
   Data<B> b;
   Data<C> c;
   // ... etc
   class Data<O extends I<O>>{
       O o;
       public void update(O o){
          this.o.update(o);
       }   
   }
}
interface I<T>{
   void update(T t);
}

class A implements I<A>{

    private String a;

    @Override
    public void update(A a) {
        this.a = a.a;
    }
}

class B extends A implements I<B>{

    private String b;

    @Override
    public void update(B b) {
       super.update(b);
       this.b = b.b;
    }
}
class C implements I<C> {public void update(C c){}}

This code cannot be compiled, because super and sub-classes trying to implements the same interface but with different type arguments.

Interface I cannot be inherited with different type arguments (A,B), anyone has a workaround to solve this?

Answer 1

No workaround are possible with such a hierarchy : in B you want to implement the i interface with two distinct generic types : A and B .
Generics are designed to bring more type safety and this possible ambiguity defeats that. From the client of the class, why update(B b) would be valid but update(A a) would be not ?
So the compiler will never accept it.

But with composition you could do :

class B  implements i<B>{
    private A a;
    private String b;

    @Override
    public void update(B b) {
       super.update(b);
       this.b = b.b;
    }
}

And now you can use the A a field if needed from the B instance.

Answer 2

With the inheritance model given below, you'll achieve a contact of i through B in A . You don't really need to implement i for class B<T> .

See the example below:

interface i<T> {


}

class A<T> implements i<T> {

}

class B<T> extends A<T> {

}

Hope it helps you to understand the structure!

Answer 3

I do not understand what is needed, but this code is compiled))

interface i<T extends i<T>> {
    void update(T t);
}

class A<Q extends A<Q>> implements i<Q> {
    private String a;

    @Override
    public void update(A a) {
        this.a = a.a;
    }
}

class B<X extends B<X>> extends A<X> implements i<X>  {
    String b;

    @Override
    public void update(X b) {
        super.update(b);
        this.b = b.b;
    }
}