what is the the natural expression of this Lambda expression and method reference in the method usage or it just don't have it?

Question

public static void main(String[] args) {
        Student[] students = new Student[4];
        students[0] = new Student("guy1", 103, 334); // height / age （randomly written）
        students[1] = new Student("guy2", 87, 534);
        students[2] = new Student("guy3", 103, 232);
        students[3] = new Student("guy4", 110, 121);

        Arrays.sort(students, Comparator.comparingInt(Student::getAge)); // method reference
        Arrays.sort(students, Comparator.comparingInt(student -> student.getAge())); // Lambda expression

// the source code of the comparingInt method
>! public static <T> Comparator<T> comparingInt(ToIntFunction<? super T> keyExtractor) {
>! Objects.requireNonNull(keyExtractor);
>! return (Comparator<T> & Serializable)
>!             (c1, c2) -> Integer.compare(keyExtractor.applyAsInt(c1), keyExtractor.applyAsInt(c2));
>!     }

I konw that I need to send the object and the method to this method, But I am confused why I can't just send the Object and method as normal parameters like Comparator.comparingInt(Student, student.get age())) (I konw it doesn't work but if I don't use Lambda expression or method reference I need to create a class in this or what?)

what I do like this is failed:

      Arrays.sort(students, Comparator.comparingInt(students,c1,getAge) kinds of , actually all can't pass IDEA.

//        Animal a = new Animal() { former version
//            @Override
//            public void run() {
//                System.out.println("dog runs");
//            }
//        };

//        Animal a = () -> {
//            System.out.println("dog runs");
//        };
//
//        a.run();

like this, I konw the the former version of the simpify version,but I can't understand how the usage and change of comparingInt sorts of. My understanding about this is that I thought they also have the Lambda in their method body,so what outside need to do is just transfer the parameter about the object and method. But why in this way? I am curious whether there is some theory or just some rules I need to remember?

My description may be not very clearly,Thank you for answering my question!

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. — Community
– Community Bot, Commented Oct 9, 2023 at 16:53

Louis Wasserman · Accepted Answer · 2023-10-09 17:43:14Z

2

You can't do anything like what you're trying to do. The closest thing is to not use comparingInt and just write (c1, c2) -> Integer.compare(c1.getAge(), c2.getAge()).

comparingInt just takes a function to run on its arguments. It does not, and cannot, take the arguments itself -- not directly, at least; you could of course write

Comparator<Student> comparator = comparingInt(Student::getAge);
comparator.compare(s1, s2);

answered Oct 9, 2023 at 17:43

Louis Wasserman

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4 Comments

Clarence Over a year ago

The compareint method actually have done the lambda thing so I can't use lambda expression to create method so on and what I should do is sending the parameter in the method reference or lambda way what inside lambda expression needed instead of sending parameter directly like what I am trying to do?

Louis Wasserman Over a year ago

If I understand correctly, the answer is that you can't do this.

Clarence Over a year ago

So I just use method refeence or Lambda like (student -> student.getAge()) to fill or stand? the parameter into the source code[ (c1, c2) -> Integer.compare(keyExtractor.applyAsInt(c1), keyExtractor.applyAsInt(c2)) ] because actually the compareint method have done the thing inside.

Louis Wasserman Over a year ago

Yes, that's correct.

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