How come Arrays.sort(stringArray, String::compareToIgnoreCase) is using compareToIgnoreCase() instead of Comparator

Need some clarification on how Arrays.sort(stringArray, String::compareToIgnoreCase) is using<br/> compareTo() whose return type is String and not Comparator which takes two arguements.

 public static void main(String[] args) {
    String[] stringArray = { "Steve", "Rick", "Aditya", "Negan", "Lucy", "Sansa", "Jon" };

    Arrays.sort(stringArray, String::compareToIgnoreCase);
    for (String str : stringArray) {
        System.out.println(str);
    }
}

A Comparator<String> is a functional interface that takes two String parameters and returns an int. Where such a Comparator is required, a lambda with the same characteristics can be used instead.

The compareToIgnoreCase method has the same characteristics, where the first parameter is the object that the method is called on, and the second parameter is the parameter to the method. The method returns an int (not a String). So it has the form of a Comparator<String>.

Because of all this, and the way functional interfaces work, one can supply a Comparator<String> object to Arrays.sort via a lambda expression that utilizes compareToIgnoreCase, like this:

Arrays.sort(stringArray, (s1, s2) -> s1.compareToIgnoreCase(s2));

The Java compiler offers a shorthand way of constructing lambdas of this form with what's called a Method Reference. The compiler will construct this same lambda (and so a Comparator<String> object) when a reference to the compareTo method is passed in via the form String::compareTo. The compiler will realize that it can call the provided method on the lambda/Comparator's first parameter (called the "receiver"), and pass the lambda/Comparator's second parameter as the single parameter that the method takes. The method will return an int, which is what any Comparator should return. And so you can do:

Arrays.sort(stringArray, String::compareToIgnoreCase);

Magic!