Java Sort with Comparable

Question

I have an ArrayList of Person objects. A Person has name, age and height. My goal is to sort this ArrayList<Person>. I have implemented Comparable<Person> and have defined compareTo() but when I try to sort it, it give me this error:

The method sort(Comparator) in the type ArrayList is not applicable for the argument ()"

The way I understand is that if you implement Comparable and then define compareTo everything else is magically done for you.

Can some one explain how to this works and why I am getting this error?

What Comparator are you passing? Can you please share some code? — Mureinik
– Mureinik, Commented Dec 19, 2016 at 15:52
use Collections.sort instead. ArrayList itself does not know how to sort without a Comparator. — njzk2
– njzk2, Commented Dec 19, 2016 at 15:54

slim · Accepted Answer · 2016-12-20 15:33:50Z

10

My guess is that your code looks like this:

ArrayList<Person> people = ...;
people.sort();

Look at the JavaDoc for ArrayList. Do you see a method public void sort() (with no parameters)? No - there is no such method.

That is the meaning of the error: The method sort(Comparator) in the type ArrayList is not applicable for the argument () -- There is a method sort(Comparator), but you have not supplied parameters that match it.

Assuming Person implements Comparable (and therefore has a compareTo() method), you can use Collections.sort(), which sorts arbitrary List<Comparable>

Collections.sort(people);

This is because Collections has a static method:

static <T extends Comparable<? super T>> void sort(List<T> list);

(it also has a sort(List<T> list, Comparator<T> comparator))

... or you can pass a comparator to List.sort(), which is quite easy with Java 8 lambdas:

people.sort((a,b) -> a.compareTo(b));

(Or, if you prefer the old style):

people.sort(new Comparator<String>() {
    @Override
    public int compare(String a, String b) {
        return a.compareTo(b);
    }
});

(Actually as of Java 8, this comparator is provided by the standard library, as Comparator.naturalOrder())

The point of comparators is that you can sort according to different criteria. For example:

people.sort((a,b) -> a.lastName().compareTo(b.lastName()));
people.sort((a,b) -> a.lastName().compareToIgnoreCase(b.lastName()));
people.sort((a,b) -> Integer.compare(a.age(),b.age()));
// etc.

... or using methods in Comparator:

people.sort(Comparator.comparing(Person::lastName));
people.sort(Comparator.comparing(Person::lastName)
        .thenComparing(Person::firstName));

edited Dec 20, 2016 at 15:33

answered Dec 19, 2016 at 16:31

slim

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2 Comments

Louis Wasserman Over a year ago

Why write (a, b) -> a.compareTo(b) when you can write Comparator.naturalOrder()?

Lev Leontev Over a year ago

You can also pass null as a comparator, which will work the same as a natural order comparator.

davidxxx · Accepted Answer · 2016-12-19 15:55:36Z

Either you use a structure which uses the Comparable interface to order its elements when you add a new element inside it :

TreeSet<Person> persons = new TreeSet<>();
Person personOne = ...
Person personTwo = ...
persons.add(personOne);
persons.add(personTwo);

Either you use a List and the Collections.sort(List<T> list) method which takes as argument the list you want to sort (there is an overload of this method but it is not relevant in your case):

List<Person> persons = new ArrayList<>();
Person personOne = ...
Person personTwo = ...
persons.add(personOne);
persons.add(personTwo);
Collections.sort(persons);

With the TreeSet, the elements are sorted as soon as added and with the List, the elements are not sorted when you add them.
Only, the call to the Collections.sort() method sorts the list.

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