Java Stream API map argument

Question

I have some misunderstanding about Java 8 static method references.

The following is a correct statement:

Stream.of("aaa", "bbb", "cccc").map(String::length);

AFAIK map requires a Function<T, R> interface as an argument with method similar to:

R apply(T t);

However, the length() method of String class doesn't accept any arguments:

public int length() {
    return value.length;
}

1) How does it correlate with apply method which needs an argument T t?

2) If I write String::someMethod doesn't it mean that someMethod should be static (since I'm calling it by class name, not by an object reference)?

Thank you!

The object instance upon which the method is to be invoked is that missing argument you are looking for. Note how you are mapping over a stream of String instances. — Thilo
– Thilo, Commented Apr 20, 2018 at 8:03
In this situation just replace a method reference with a lambda expression for a better understanding. — Oleksandr Pyrohov
– Oleksandr Pyrohov, Commented Apr 20, 2018 at 8:36

Eran · Accepted Answer · 2018-04-20 08:03:04Z

6

No, String::someMethod doesn't mean the method has to be static. It could be either a static method, or an instance method that would be executed on some String instance. That instance would serve as an implicit argument of the single method of the functional interface implemented by that method reference.

Therefore String::length does have a single argument - the String instance on which the length method will be called.

String::length is equivalent to the lambda expression (String s) -> s.length() (or just s -> s.length()).

When you write Stream.of("aaa", "bbb", "cccc").map(String::length), the length() method would be executed for each element of your Stream (assuming you add some terminal operation that causes map to be evaluated on these elements) which would transform your Stream<String> to a Stream<Integer>.

answered Apr 20, 2018 at 8:03

Eran

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Oleksandr Pyrohov · Accepted Answer · 2018-04-26 22:22:19Z

3

When you expand your lambdas/method references to anonymous inner class, it makes things easier to understand. Your statement translates to something like this when you replace method references with anonymous class:

Stream.of("aaa", "bbb", "cccc").map(new Function<String, Integer>() {
    @Override
    public Integer apply(String s) {
        return s.length();
    }
});

As you can see here input is the String and it returns string length which is a Integer. And it adheres to the Function functional Interface definition that you mentioned.

edited Apr 26, 2018 at 22:22

Oleksandr Pyrohov

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answered Apr 20, 2018 at 8:02

pvpkiran

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