Invoke Arrays.sort( array, comparator ) using generics problem

Question

I'm trying to create a method which in turn should call Arrays.sort method and I'm trying this method to be generic itself. The problem comes when I try to create a concrete Comparator but I always get one exception like:

Generics.java:15: cannot find symbol
symbol  : method sort(java.lang.Comparable[],java.util.Comparator<T>)
location: class java.util.Arrays
                Arrays.sort(array,  c);
                      ^
1 error

I workaround it by casting the value, but I would really like to know how to code it properly.

Here's my ( relevant ) code:

import java.util.*;
abstract class Generics<T1,T2> { 

  public abstract  void run( T1 a , T2 b );

  public static <T> void sort( Comparable[] array, final Generics<T,T> g  ) {

    Comparator<T> c = new Comparator<T>() {
      public int compare(T o1, T o2) {
        g.run(o1,o2);
        return 1;
      }
    };
    Arrays.sort(array, 
      /* how can I get rid of this cast? */ 
      (Comparator<? super Comparable>)
      /* and use only */ c);
  }

  public static void main( String ... args ) { 
    Generics.sort( args, new Generics<String,String>() { 
      public void run( String a, String b ) { 
        System.out.println( "Irrelevant");
      }
    });
  }
}

Also note that your Comparator implementation (with return 1) does not fulfill the contract of the interface (antisymmetry, reflexivity). I suppose this is only for testing purposes. — Paŭlo Ebermann
– Paŭlo Ebermann, Commented May 24, 2011 at 23:08
Yeap, that part is not relevant to the problem. Thanks again. — OscarRyz
– OscarRyz, Commented May 24, 2011 at 23:09

Paŭlo Ebermann · Accepted Answer · 2011-05-24 23:02:58Z

6

Your method should not pass a Comparable[], but a T array, if you want to use a Comparator<T>.

Thus define it this way:

public static <T extends Comparable<T>> void sort( T[] array, final Generics<T,T> g  ) {
    ...
}

But why do you want to use both Comparable and Comparator? Normally the idea is to either use the Comparator or the default order on the element (i.e. Comparable), but not both at once.

answered May 24, 2011 at 23:02

Paŭlo Ebermann

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2 Comments

OscarRyz Over a year ago

I struggle with this for ... hours! :) :) Thans a lot Paulo!

Don Roby Over a year ago

+1: But as it makes little sense to use both Comparable and Comparator, you don't need the "extends Comparable<T>".

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