Understanding QuickSort Syntax

Question

I'm confused by this syntax because at first it appears as though the function should return <E> but then I see void . What is the purpose of the <E> prior to the void ?

Also, I'm familiar with bounding and generics, but I've never actually seen something bounded this way. What does this syntax mean Comparator<? super E> Comparator<? super E> ?

Here is a sample function:

private <E> void sort(E[] array, Comparator<? super E>  cmp) {
   qsort(array, 0, array.length - 1, cmp);
}

Answer 1

The first <E> is not a type - it is a type constraint.

Bear in mind that Java implements generics via type erasure - this means that the runtime type signature of this method is

private void sort(Object[] array, Comparator cmp)

(by removing everything in between the <> s) so your method has return type void .

What <E> does is to say that the types of the input array and comparator are related: the comparator needs to be able to compare 'things' of type E , but it actually doesn't have to only handle things of the exact type E .

This is what <? super E> <? super E> does: for example, you could have a Comparator<CharSequence> , and then use that to sort String[] , since String is a subclass of CharSequence .

Answer 2

at first it appears as though the function should return <E> but then I see void

The <E> part there is a syntax for supplying generic methods their type argument. The actual return type is void .

What does this syntax mean Comparator<? super E> Comparator<? super E>

This means that the comparator may be for E or for any of its superclasses.

Answer 3

Comparator<? super E>  

? is the wildcard character and the lower bound for this is E.

for example:

private <E> void sort(E[] array, Comparator<? super E>  cmp)

if you are passing

qsort(array, 0, array.length - 1, cmp);

if array is Number[] then lower bound of "?" is set to java.lang.Number ie "?" can be anything which is a super class of Number.