Java - sorting using Collections.sort()

Question

I have to write a method to sort both Integers and Doubles.

public ArrayList<Number> sortDescending(ArrayList<Number> al){
    Comparator<Number> c=Collections.reverseOrder();
    Collections.sort(al,c);
    return al;
}

public ArrayList<Number> sortAscending(ArrayList<Number> al){
    Collections.sort(al);
    return al;
}

The problem is that in sortAscending, the following error occurs:

Bound mismatch: The generic method sort(List) of type Collections is not applicable for the arguments (ArrayList). The inferred type Number is not a valid substitute for the bounded parameter < T extends Comparable < ? super T>>

Bohemian · Accepted Answer · 2012-07-16 14:36:02Z

9

You need to use a generic upper bound of Number intersecting with Comparable<T>:

public <T extends Number & Comparable<T>> ArrayList<T> sortDescending(ArrayList<T> al){
    Comparator<T> c=Collections.reverseOrder();
    Collections.sort(al,c);
    return al;
}

public <T extends Number & Comparable<T>> ArrayList<T> sortAscending(ArrayList<T> al){
    Collections.sort(al);
    return al;
}

All JDK Numbers (eg Float, Integer etc) match this typing.

For the uninitiated, the syntax <T extends A & B> is the way you bound T to both A and B.

FYI, there is no syntax for "or" logic (nor would it make sense if you think about it)

answered Jul 16, 2012 at 14:36

Bohemian♦

427k103 gold badges603 silver badges750 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Matt Over a year ago

FYI: There's no reason to have a return type on either of your methods, as the List is sorted in place.

Bohemian Over a year ago

@Matt Rather than "no reason", there are several reasons to have return types: A) To illustrate the implications of typing the method, B) That's what the OP had and C) Returning objects from "non-getter" methods is done when implementing the fluent API pattern

Oskar Kjellin · Accepted Answer · 2012-07-16 14:35:47Z

4

You are getting the error because number does not implement Comparable<Number>. You need to add a generic contstraint so that it extends both number and Comparable. In this case:

public <T extends Number & Comparable<T>> ArrayList<T> sortAscending(ArrayList<T> al){
    Collections.sort(al);
    return al;
}

answered Jul 16, 2012 at 14:35

Oskar Kjellin

22k10 gold badges57 silver badges95 bronze badges

Collectives™ on Stack Overflow

Java - sorting using Collections.sort()

2 Answers 2

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related