Java: Different outputs when add/remove short and integer elements in a Set

Question

I wasn\'t sure on how to ask this question. But, what is the different between these 2 lines of code?

Set a = new HashSet();
f         


        

Answer 1

The type of the expression i - 1 is int because all operands in an integer arithmetic expression are widened to at least int. Set<Short> has add(Short) and remove(Object) so there's no casting/autoboxing needed on the remove call. Therefore you are trying to remove Integers from a set of Shorts.

Note that for this reason it almost never makes sense to declare a Set<Number>:

final Set<Number> ns = new HashSet<>();
final short s = 1;
ns.add(s);
ns.add(s+0);
ns.add(s+0L);
System.out.println(ns); // prints [1, 1, 1]

As a bonus round, if you change the set implementation to TreeSet, the magic disappears and it throws a ClassCastException, giving away the trick.

Deep down, this issue has to do with the fact that equality is a symmetric relation, which must not distinguish the right hand side from the left hand side. These semantics are impossible to achieve with Java's single-dispatch methods.

Answer 2

The first code snippet removes from a HashSet of Integers all the numbers except for 99 because 98 is last number it removes.

The second code snippet is trying to remove an Integer from a HashSet of Shorts, consequently, it does not remove any element.

On the first code snippet, in the statement add(i) the int is automatically converted to an Integer.

On the second code snippet, if you had done the following:

Set<Short> a = new HashSet<Short>();
for (Short i = 0; i < 100; i++) {
     a.add(i);
     a.remove(i);
}

It would remove all the elements, since you add and remove a Short. However, because you are trying to remove i - 1, it will convert i - 1 to an Integer. Hence, trying to remove an Integer from a HashSet of Shorts, which actually leads to no number being removed.

Answer 3

You can't remove Integer from Set of shorts without casting.