0 
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?

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10
  • What compiler error message do you get? Commented Apr 7, 2016 at 14:08
  • 3
    because you inherit from A, which has a generic in it´s class signature. If you leave it out A will be inherited as rawtype.(how should the compiler automaticly decide for you that the generic E for class B has to be used as the generic for A aswell)? Commented Apr 7, 2016 at 14:09
  • 1
    How does the compiler know that A<E> doesn't extend B<String>? The type parameter might be used for something quite different in the subclass. Commented Apr 7, 2016 at 14:10
  • 2
    for the second question, super(E), what is E, it´s not a variable just a type, so the compiler tells you so (what should it pass?) Commented Apr 7, 2016 at 14:10
  • 2
    @LetDoit you can always pass (E) null. Commented Apr 7, 2016 at 14:17

1 Answer 1

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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; i.e. 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);
    }
}
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2 Comments

How to pass to parent's constructor without child constructor's parameter? super(3); also not works.
super(3) would work if and only if 3 is an instance of the generic parameter E. But since this parameter has no bounds, it could be any type, and so the compiler correctly says this is not consistent. (If you tried to construct a B<String>, which is thus a subclass of A<String>, you'd be trying to pass an int into a constructor that takes a String). In general with unbounded parameters, you'll need to have the caller provide the constructor parameter for this reason.

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