I am trying to convert a Set to an Array.
Set<String> s = new HashSet<String>(Arrays.asList("mango","guava","apple"));
String[] a = s.toArray(new String[0]);
for(String x:a)
System.out.println(x);
And it works fine. But I don't understand the significance of new String[0] in String[] a = s.toArray(new String[0]);.
I mean initially I was trying String[] a = c.toArray();, but it wan't working. Why is the need for new String[0].
It is the array into which the elements of the Set are to be stored, if it is big enough; otherwise, a new array of the same runtime type is allocated for this purpose.
Object[] toArray(), returns an Object[] which cannot be cast to String[] or any other type array.
T[] toArray(T[] a) , returns an array containing all of the elements in this set; the runtime type of the returned array is that of the specified array. If the set fits in the specified array, it is returned therein. Otherwise, a new array is allocated with the runtime type of the specified array and the size of this set.
If you go through the implementing code (I'm posting the code from OpenJDK) , it will be clear for you :
public <T> T[] toArray(T[] a) {
if (a.length < size)
// Make a new array of a's runtime type, but my contents:
return (T[]) Arrays.copyOf(elementData, size, a.getClass());
System.arraycopy(elementData, 0, a, 0, size);
if (a.length > size)
a[size] = null;
return a;
}
The parameter is a result of one of the many well-known limitations in the Java generics system. Basically, the parameter is needed in order to be able to return an array of the correct type.
