Question

I thought that null is allowed for a Set.
So why does the following code:

SortedSet<Integer> set = new TreeSet<Integer>();  
set.add(null);  
set.add(1);  //--->Line indicated by exception  

Gives the following exception?

Exception in thread “main” java.lang.NullPointerException at
java.lang.Integer.compareTo(Unknown Source) at
java.lang.Integer.compareTo(Unknown Source) at
java.util.TreeMap.put(Unknown Source) at
java.util.TreeSet.add(Unknown Source)

Answer 1

Yes, you can. But you will have to provide your own Comparator to handle the case when null is compared to any other contents of your set. With natural ordering applied, Java objects do not know how to compare themselves to null. Inversely, null doesn’t know how to compare itself with any object as you cannot call null.compareTo(object).

An example implementation of such a “null-safe” Comparator can be found in the apache commons-collections library. Check out the NullComparator. You could use it as such:

// Unfortunately no support for Java generics yet, in commons-collections
@SuppressWarnings("unchecked")
SortedSet<Integer> set = new TreeSet<Integer>(new NullComparator());  
set.add(null);  
set.add(1);