Re: I need to retrieve an Object[] of all keys in java.util.Hashtable

Lew <>
Tue, 13 Feb 2007 16:23:20 -0500
<> wrote:


public abstract class ArrayFunctionality {

     * Construct {@link java.lang.Object} array of keys from {@link
     * @param h {@link java.util.Hashtable}
     * @return array {@link java.lang.String}
     * @throws java.lang.IndexOutOfBoundsException Exception thrown if
initial {@link java.lang.Object} array paramater cannot be indexed
    public static Object[] arrayKeys(Hashtable<Object, Object> h)
throws IndexOutOfBoundsException {

IndexOutOfBoundsException is not a checked exception, and isn't thrown anyway.

        Vector<Object> v = new Vector<Object>();
        Enumeration keys = h.keys();
        while (keys.hasMoreElements()) v.add(keys.nextElement());
        return v.toArray();


Once you've fixed your errors, you might consider using HashMap instead of
Hashtable, Iterator instead of Enumeration (which, by the way, you did not
genericize), and ArrayList instead of Vector.

Another thing you might consider is using the generic types more fully than
merely <Object>. For example, you might use something like this untried code:

   public abstract class AFunc<T>
     public static Object [] arrayKeys( Map<T, ?> map )
       Set<T> keys = map.keySet();
       return keys.toArray();

You can do this without a method as a one-liner. Given a Map<T, ?> map:

     Object [] keys = map.keySet().toArray();

This might make the effort of writing a method seem excessive.

There are tricks to play with a Class<T> instance variable to let you generate
a type T array, something like this untried code using a Class<T> variable

   private final T [] ARR =
         java.lang.reflect.Array.newInstance( clazz, 0 );
   T [] keys = map.keySet().toArray( ARR );

Of course, if you use the one-liner then you probably have type information
for the array without resorting to reflection.

There are folks more facile with generics than I, so I might not have
expressed the idioms as well as they would have.

- Lew

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