Question on Generics Wild Card

[Thiyagarajan Kamala]

List<? extends Animal> genericExtends = new ArrayList<Animal>();
This is not allowing me to add anything to it then why allow to instantiate a list .

[Balraj Momi]

List<? extends Animal> genericExtends

"Says this reference can be initialized with Animal or anything that extends Animal e.g. CAT, DOG, etc. This was about initialization.

But once initialization is done, compiler has no idea what this reference is initialized to, It can be anyone from CAT, DOG, or Animal. Since It is not sure that what object type is this, compiler gives compilation error.

[Thiyagarajan Kamala]

Why not let me add anything that is Animal or sub type of Animal I have already told the compiler it is referring to Animal.

Apprecicate your reply.

[Balraj Momi]

My Friend, object creation happens at Runtime not at Compile time.

Read this Generics Super and Extends

you will get full insight what I was trying to say. Go through this, then come back If you have any question on this.

[Ankit Garg]

Thiyagarajan Kamala wrote:List<? extends Animal> genericExtends = new ArrayList<Animal>();
This is not allowing me to add anything to it then why allow to instantiate a list .

There is use of this indeed. Suppose you have this structure

class Animal {
    public void makeSound() {
        System.out.println("BlahBlah");
    }
}
class Dog extends Animal {
    public void makeSound() {
        System.out.println("Woof");
    }
}
class Main {
    public static void makeSounds(List<? extends Animal> list) {
        for(Animal a: list) {
            a.makeSound();
        }
    }
    public static void main(String[] args) {
        List<Dog> dogs = new ArrayList<Dog>();
        //add elements to the list
        makeSounds(dogs);
        List<Animal> animals = new ArrayList<Animal>();
        //add elements to the list
        makeSounds(animals);
    }
}

So the thing to say is, you cannot add any element to such a list, but you can definitely get elements out of it. So it can be useful if the list is already populated...

[Monu Tripathi]

Remember:

Generics was introduced for type safety, so that you can detect and prevent "possible" runtime errors at compile time.

Also, compiler adds required casts whenever necessary so, no explicit casting is required.

List<? extends Animal> genericExtends = new ArrayList<Animal>();

When you write this statement, you are saying that genericExtends can be assigned to any List-type which holds Objects of type Animal or its sub classes. So, the only thing you are guaranteeing to the compiler is that the UPPER BOUND of the type in the list will be Animal.

Compiler cant allow you to add any object to this list, because it does NOT know the actual type of List that you assign at Run time. Add operation will have the risk of ClassCastException. For e.g.:

// Type definition
class A {}
class B extends A {}
class C extends B {}

public class GenericsTest {
	public static void main(String[] args) throws Exception{
		// Only thing that the compiler knows here is that
		// the upper bound on TYPE is "B"
		// we cannot add anything to lst though.
		// the reason why we cannot add anything to it is:
		// the fact that the compiler does NOT know anything about the 
		// type of list we are assigning at run time
		// for e.g if we assign a List<B> to the reference and
		// try to add an "A" instance, we will get a ClassCastException
		List<? extends B> lst = new ArrayList<B>();
//Remember: compiler does not know what the Run time type is, not even that it is B
		// lst.add(new A());// compiler error. 
		// lst.add(new B());// compiler error
		// lst.add(new C());// compiler error

Compiler will allow you to get/read an object from this list, because it knows that the object returned will be atleast of type B or Animal(in your case). So the cast will never fail.

B obj = lst.get(0); // this is allowed, since any object will atleast be of type B.

Hope that helps...

Read this for more info:
Sun Tuts: Wild Cards
Generics : What is a wild card