Why does the constructor of TreeSet<E> accept the parameter super of E instead of extend of E

Question

I'm reading the java source code and I've found things as below:
http://www.java2s.com/Code/JavaAPI/java.util/newTreeSetEComparatorsuperEc.htm

I don't understand why the parameter of this constructor is <? super E>.

As my understanding, it should be <? extend E> instead of <? super E> because if E is comparable, the children of E must be comparable whereas the parents of E may be not.

Answer 1

Let's consider three classes: Drink, Juice extends Drink, and OrangeJuice extends Juice.

If I want a TreeSet<Juice>, I need a comparator that will compare any two juices. Of course a Comparator<Juice> will do this.

A Comparator<Drink> will also do this, because it is able to compare any two drinks, and thus any two juices.

A Comparator<OrangeJuice> will not do this. If I wanted to add an AppleJuice to my set of juices, that is not compatible with this comparator as it can only compare orange juices.

Answer 2

You create a sorted set of bananas. To do that, you need to be able to compare the bananas. You can thus use a comparator of bananas to do that. But if you have a comparator that can sort any kind of fruit (including bananas), then that'll work too. So you can also use a Comparator<Fruit> or even a Comparator<Food> to sort your bananas.

That's why a Comparator<? super E> is used.

If it was Comparator<? extends E>, you would be able to create a set of bananas with a Comparator<DriedBanana>. But that wouldn't work: a comparator of dried bananas can only compare dried bananas. Not all kinds of bananas.

For more information, also read What is PECS (Producer Extends Consumer Super)?