bounded type parameters

 public <U extends Number> void inspect(U u){
        System.out.println("T: " + t.getClass().getName());
        System.out.println("U: " + u.getClass().getName());

    public static void main(String[] args) {
        Box<Integer> integerBox = new Box<Integer>();
        integerBox.set(new Integer(10));
        integerBox.inspect("some text"); // error: this is still String!

generic methods

public class Util {
    // Generic static method
    public static <K, V> boolean compare(Pair<K, V> p1, Pair<K, V> p2) {
        return p1.getKey().equals(p2.getKey()) &&
boolean same = Util.<Integer, String>compare(p1, p2);

generic type example?

Box<Integer> integerBox;
public class Box<T> {
    // T stands for "Type"
    private T t;

    public void set(T t) { this.t = t; }
    public T get() { return t; }

The Diamond

In Java SE 7 and later, you can replace the type arguments required to invoke the constructor of a generic class with an empty set of type arguments (<>) as long as the compiler can determine, or infer, the type arguments from the context. This pair of angle brackets, <>, is informally called the diamond. For example, you can create an instance of Box<Integer> with the following statement:

Box<Integer> integerBox = new Box<>();


  • Type as parameters

why use?

  • Stronger type checks at compile time.
  • Enabling programmers to implement generic algorithms.
  • Elimination of casts.The following code snippet without generics requires casting:
    List list = new ArrayList();
    String s = (String) list.get(0);

    When re-written to use generics, the code does not require casting:

    List<String> list = new ArrayList<String>();
    String s = list.get(0);   // no cast

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