Inheritance using Generics in Java

Question

class A<E>{

}
class B<E> extends A{

}

public class WildInDeclare<E>{
    public static void main(String[] args){
        A<Integer> obj = new B<Integer>();
    }
}

When I compile the above program I got this error.

Note: WildInDeclare.java uses unchecked or unsafe operations.
Note: Recompile with -Xlint:unchecked for details.

After that I just replace class B<E> extends A{ to class B<E> extends A<E>{ it works fine can you explain why it required to again write A<E> During inheritance with code Example Please.

The second Problem with Inheritance I faced. I update the above code with the follow code.

class A<E>{
    public A(E o){

    }
}
class B<E> extends A{
    public B(){
        super(E);
    }
}

public class WildInDeclare<E>{
    public static void main(String[] args){
        A<Integer> obj = new B<Integer>();
    }
}

But it is not compiling why? I got the following error.

WildInDeclare.java:8: error: cannot find symbol
                super(E);
                      ^
  symbol:   variable E
  location: class B<E>
  where E is a type-variable:
    E extends Object declared in class B
Note: WildInDeclare.java uses unchecked or unsafe operations.
Note: Recompile with -Xlint:unchecked for details.
1 error

How I can pass Parameter to Parent class?

Answer 1

can you explain why it required to again write A

Because A is a generic class that takes one type parameter. Ideally, you wouldn't be able to refer to the raw type A , you'd only be able to refer to A<String> or A<Integer> or A<Map<K, List<O>> etc. Java lets you use the raw type for backwards compatibility, but in principle it should be compulsory to provide the parameter.

When you're saying that class B extends A , you still need to say what kind of A it extends; ie what the generic parameter is. (This does not need to be the same as B 's generic parameter - you could for example define B<E> extends A<String> and that would be consistent.)

the following code...is not compiling. Why?

That code is just syntactically invalid. You've defined A 's constructor to take an object of its generic parameter type. So when you call super() from B 's constructor, you need to pass in such an object. E is not an object - it's just the name for B's generic parameter. So the compiler is right to say "cannot find symbol", there is nothing called E in scope.

If you wanted B to take an input parameter that it just passes to the superclass constructor, it would look something like this:

class B<E> extends A<E> {
    public B(E e) { // this parameter could be called anything, doesn't have to be e
        super(e);
    }
}