使用比较器对流排序的方法

Question

I don't understand why code1 works but code2 doesn't compile. Please explain.

//Code1:
    Stream<String> s = Stream.of("AA", "BB");
    s.sorted(Comparator.reverseOrder())
            .forEach(System.out::print);

//Code2:
    Stream<String> s = Stream.of("AA", "BB");
    s.sorted(Comparator::reverseOrder)
            .forEach(System.out::print);

The difference between the two is code1 uses Comparator.reverseOrder() while code2 uses Comparator::reverseOrder

Answer 1

Because the first example is a factory-method so when you inspect it, you see that you get a comparator back.

But the second one is a method-reference which you could write like this:

Stream<String> s = Stream.of("AA", "BB");
s.sorted(() -> Comparator.reverseOrder()) // no semantic difference!
    .forEach(System.out::print);

But it has a whole different meaning because this time you are given Stream#sorted() a Supplier<Comparator<?>> but it just needs a Comparator<?>

Small Sidenote: Don't store streams in variables, use them directly. So i would suggest you just write:

Stream.of("AA", "BB")
    .sorted(Comparator.reverseOrder())
    .forEach(System.out::print);

Answer 2

The error message from the compiler should tell you that.

sorted() expects a Comparator instance. Comparator.reverseOrder() returns a Comparator instance. So that works fine.

Comparator::reverseOrder is a method reference to the reverseOrder() method of Comparator. So your code basically says: each time you need to compare two strings, pass them as argument to Comparator.reverseOrder to compare them. But that can't possibly work. This method takes nothing as argument, and returns a Comparator. So it doesn't match the signature of a Comparator<String> , which is supposed to take two Strings as argument, and return an integer.

If you had a method such as

class Foo
    public static int compareStrings(String s1, String s2) {
        ...
    }
}

Then you could use

sorted((s1, s2) -> Foo.compareStrings(s1, s2))

which you can transform, using a method reference, to

sorted(Foo::compareStrings)

Because compareStrings, just like the unique abstract method of Comparator<String> , takes two Strings as argument and returns an int.