Wsrtjtyk

I'm learning about lambda expressions. I don't understand how a comparator is returned from method reference.

I want to sort a list of persons by age.

To do that, I have a method to find the age difference:

public int ageDifference(final Person other) {

    return age - other.age;

}

The sorted method needs as parameter a Comparator

Stream<T> sorted(Comparator<? super T> comparator);

My lambda expression is:

people.stream()

.sorted(Person::ageDifference)

.collect(toList());

How Person::ageDifference is trasformed in a Comparator<Person>?

My complete example:

public class Person {



private final String name;

private final int age;



public Person(final String theName, final int theAge) {

    name = theName;

    age = theAge;

}



public String getName() {

    return name;

}



public int getAge() {

    return age;

}



public int ageDifference(final Person other) {

    return age - other.age;

}



public String toString() {

    return String.format("%s - %d", name, age);

}



public static void main (String args ){

    final List<Person> people = Arrays.asList(

            new Person("John", 10),

            new Person("Greg", 30),

            new Person("Sara", 20),

            new Person("Jane", 15));



    List<Person> ascendingAge =

            people.stream()

                    .sorted(Person::ageDifference)

                    .collect(toList());



    System.out.println(ascendingAge);

}

}

Output: [John - 10, Jane - 15, Sara - 20, Greg - 30]

I guess your main confusion is this:

Comparator<T> represents a method that takes two parameters of type T and returns a int. My ageDifference method accepts only one Person parameter. How can that become a Comparator<Person>?

Note that ageDifference is an instance method. To call it, not only do you need the parameters, you also need an instance of Person. In this case, you need 2 Persons to call the method - one on which you call ageDifference, and the other one you pass as a parameter:

me.ageDifference(someoneElse)

^                      ^

|                      |

        Two people!

Isn't this just like a static method that accepts two parameters?

Therefore, Java is smart enough to know that you need two people to call Person::ageDifference, so the method reference is treated as having two parameters.

In general, an instance method of a class T accepting parameters P1, P2 ... Pn and returning type R can be treated as a static method accepting parameters T, P1, P2 ... Pn and returning R.