Answer by Varkhan for How can I create a generic array in Java?

I have to ask a question in return: is your GenSet"checked" or "unchecked"?What does that mean?

• Checked: strong typing. GenSet knows explicitly what type of objects it contains (i.e. its constructor was explicitly called with a Class<E> argument, and methods will throw an exception when they are passed arguments that are not of type E. See Collections.checkedCollection.

  -> in that case, you should write:

  public class GenSet<E> {    private E[] a;    public GenSet(Class<E> c, int s) {        // Use Array native method to create array        // of a type only known at run time        @SuppressWarnings("unchecked")        final E[] a = (E[]) Array.newInstance(c, s);        this.a = a;    }    E get(int i) {        return a[i];    }}

• Unchecked: weak typing. No type checking is actually done on any of the objects passed as argument.

  -> in that case, you should write

  public class GenSet<E> {    private Object[] a;    public GenSet(int s) {        a = new Object[s];    }    E get(int i) {        @SuppressWarnings("unchecked")        final E e = (E) a[i];        return e;    }}

  Note that the component type of the array should be the erasure of the type parameter:

  public class GenSet<E extends Foo> { // E has an upper bound of Foo    private Foo[] a; // E erases to Foo, so use Foo[]    public GenSet(int s) {        a = new Foo[s];    }    ...}

All of this results from a known, and deliberate, weakness of generics in Java: it was implemented using erasure, so "generic" classes don't know what type argument they were created with at run time, and therefore can not provide type-safety unless some explicit mechanism (type-checking) is implemented.