Sorting generic data structure array

Question

I have a generic class that contains an array. That array can be any data type(Double,Integer,String...) and I am trying to learn how could I sort this array just after constructed. I've wrote a sorting method for sorting the array after constructed. But, java inhibits sorting generic array with mysort method due to mismatch.

public class SortingTest<E> {

    private E[] array;
    //constructor which creates an object and sorts its array via mysort method.
    public SortingTest(E[] N){
        this.array = N;
        mysort(this.array);
    }
    //sorts array.
    private static<T extends Comparable<T>> void mysort(T[] arr){
        T temp = null;
        //Bubble sort increasing order.
        for(int j=0; j<arr.length; j++){
            for(int i=1; i<arr.length; i++){
                if(arr[i-1].compareTo(arr[i]) > 0 ){
                    temp = arr[i-1];
                    arr[i-1] = arr[i];
                    arr[i] = temp;

                }
            }
        }
    }
}

How can I overcome this problem? Any help is appreciated.

Think about what you are passing to mysort(..) in your constructor. What type is it? Does it match the bounds set by the method. — Sotirios Delimanolis
– Sotirios Delimanolis, Commented Jun 5, 2014 at 17:36
It can be any type as I said. The problem , which arise from didn't known implementation of Comparable interface, is solved by @Luiggi Mendoza Thank you very much for help. — Sixie
– Sixie, Commented Jun 5, 2014 at 17:41
As an aside, note that generics and arrays don't play well together. You'll have a much better time working with Lists and the rest of the Collections Framework. — dimo414
– dimo414, Commented Jun 5, 2014 at 17:41

Luiggi Mendoza · Accepted Answer · 2014-06-05 17:47:12Z

3

mysort method requires generic element T implements Comparable<T>, but generic E doesn't declare that implements Comparable<E>. Change the declaration to:

public class SortingTest<E extends Comparable<? super E>> {
    //rest of your code...
}

edited Jun 5, 2014 at 17:47

answered Jun 5, 2014 at 17:36

Luiggi Mendoza

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3 Comments

pscuderi Over a year ago

I think it would be better if you did "E extends Comparable<? super E>" to cover the case where the parent class implements Comparable. Sorry if that's nit-picky.

Sixie Over a year ago

What is the difference between them? Does using super keyword make any advantages? @pscuderi

Luiggi Mendoza Over a year ago

@Sixie it means that if the generic class E extends from another class SE and SE is the class that implements Comparator<SE>, then E won't need to implement Comparator<E> itself because already implements Comparator<SE>.

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