Generics

[Minu Sam]

I am having hard time understanding generics.
Could someone explain why line 2 compilation fails.??

public class TestingGeneric {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		addAnimal(new Animal());
		addAnimal(new Dog());
		
	}
	
	
	static <T extends Animal> void addAnimal(T t)
	{
		
	}

	//Doesnt compile -Line 2
	static <P super Animal> void addAgainWithSuper(P p)
	{
		
	}

}
Assuming Dog extends Animal

[Henry Wong]

Minu Sam wrote:I am having hard time understanding generics.
Could someone explain why line 2 compilation fails.??

public class TestingGeneric {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		addAnimal(new Animal());
		addAnimal(new Dog());
		
	}
	
	
	static <T extends Animal> void addAnimal(T t)
	{
		
	}

	//Doesnt compile -Line 2
	static <P super Animal> void addAgainWithSuper(P p)
	{
		
	}

}
Assuming Dog extends Animal

Simply, lower bounds are not supported with generic methods. That's how it is defined.

Henry

[Minu Sam]

Henry,

Appreciate your response, however can you please explain what you mean by lower bounds?I still dont understand

[Henry Wong]

Minu Sam wrote:
Appreciate your response, however can you please explain what you mean by lower bounds?I still dont understand

Java generics allow you to bound the type. You can specify an upper bound via <T extends UpperBound>, or you can specify a lower bound via <T super LowerBound>. With generic methods, only upper bounds are supported.

Henry

[Helen Ma]

//Doesnt compile -Line 2   
    static <P super Animal> void addAgainWithSuper(P p)   
    {   
           
    }  

This wont' compile because p can be Animal, Animal's super type or Object.
What happens if a programmer does this inside addAgainWithSuper method :
List<Animal> l = new ArrayList<Animal>();
l.add(p);

?
What happens if p is an object? The list l cannot add p to it.

Or what happens if a program writes this:
Animal a = p; //where p is an object.
It will have type mismatch compilation error.

Since the programmer can write code in the method that may violation compilation rules or not type safe, the compiler won't let the addAgainWithSuper to be compiled as p may be Animal, Animal's super type or even object. During compilation time, the compiler does not know what type of p the programmer will pass as the argument.

Correct me if I am wrong.