Generics

• Generics is a way to enforce ONLY compile-time type safety.
• All the type information is not present at run-time. The compiler strips out type information from the bytecode using a process called type erasure.
• WHY Type erasure ? To ensure backward compatibility with legacy code.
• This compile-time safety is broken when generic and non-generic legacy code are mixed up.

See below:

       private void bar() {
		List<Integer> li = new ArrayList<Integer>();
		li.add(new Integer(1));
		li.add(new Integer(2));
		foo(li);
		for(Integer i : li) {
			System.out.println(i.intValue()); // This will fail with a ClassCastException.
		}
	}
	
	private static void foo(List l) {
		l.add(new Integer(3));
		l.add(new String("4")); //Compiler ALLOWS this !  However, warning will be generated.
	}

• Watch out when autoboxing is involved with legacy code.

List l = new ArrayList();
l.add(123); //Auto-boxing happens.
int i = l.get(0); //Compile-time error. Autounboxing cant work because get() returns Object and not Integer.

Polymorphism and Generics

List<String> list = new ArrayList<String>();

Polymorphism only applies to the base type i.e list can be declared as arraylist You CANNOT do this:

 List<Animal> obj = new ArrayList<Dog>(); //NOT POSSIBLE

WHY ?

To prevent scenarios where you cannot add say, a Cat object to a Dog List. If the above conversion were possible it will be possible to do so. See below:

//NOTE : This is not possible actually, because the compiler prevents it.
public void foo() {
   List<Dog> dList = new ArrayList<Dog>();
   addAnimal(dList); //Compiler flags an error here. a Dog list cannot be assigned to an Animal list
}

private void addAnimal(List<Animal> aList) {
  aList.add(new Cat());
}

However, the SAME thing is possible with Arrays

public void foo() {
   Dog[] dA = new Dog[]{};
   addAnimal(dList); 
}

private void addAnimal(Animal[] aa) {
  aa[0] = new Cat(); //This will cause a runtime ArrayStoreException
}

• The reason why it such polymorphism is possible with Arrays but not with collections is because of Type Erasure.
• Since there is no type information at run-time, JVM cannot raise an exception.
• This will be exactly the same problem when type-safe collections are mixed with non-type safe ones.
• So, the compiler will prevent such polymorphic assignments when we are dealing with type-safe collections.

• Q : How to use generic collections polymorphically then ?
• A : Use wildcards ( ? )

• See example below. This means all Lists of Dogs, Cats, Elephants etc can be passed to the move method.
• However the compiler ensures that the list cannot be modified to avoid wrong animals being inserted into the wrong lists.
• How ? Any method in the List interface which accepts a generic parameters like add(E) or bar(E) will not be allowed.

       public static void foo() {
		List<Dog> ld = new ArrayList<Dog>();
		foo(ld);
	}
	
	private static void move(List<? extends Animal> wildL) {
		for(Animal a : wildL) {
			a.move();
		}
	}

• If you do want to modify the list, the wildcard declaration must be made safe by using the keyword super.
• In the below example, it means that all Lists containing objects of superclasses of Dog can be passed to the foo method.
• So foo is safe to add Dog objects to the passed list, since the passed list is guaranteed to be a list of some objects which are above Dog. So type-safety is maintained.

       public static void bar() {
		List<Dog> ld = new ArrayList<Dog>();
		foo(ld);
	}
	
	private static void foo(List<? super Dog> dogList) {
              dogList.add(new Dog());
              dogList.add(new Animal()); //WONT WORK. 
	}

• NOTE VERY IMPORTANT: Wildcards specify what kind of collections the method can accept. NOT what can be added to the collection

• Only using the wildcard - like below, means that any list can be passed to it. But of course nothing can be added to the list.

        void foo(List<?> anyList)

• List<?> list is different from List<Object> objList ! The second will take only lists which contain Objects, whereas the former will take anything.
• It follows that List<? extends Object> is the same as List<?>