Java8 method reference used as Function object to combine functions

Question

Is there a way in Java8 to use a method reference as a Function object to use its methods, something like:

Stream.of(\"ciao\", \"hola\", \"hello         


        

Answer 1

You can write a static method to do this:

import java.util.function.*;

class Test {
    public static void main(String[] args) {
        Function<String, Integer> function = combine(String::length, n -> n * 2);
        System.out.println(function.apply("foo"));
    }

    public static <T1, T2, T3> Function<T1, T3> combine(
        Function<T1, T2> first,
        Function<T2, T3> second) {
        return first.andThen(second);
    }
}

You could then put it in a utility class and import it statically.

Alternatively, create a simpler static method which just returns the function it's given, for the sake of the compiler knowing what you're doing:

import java.util.function.*;

class Test {
    public static void main(String[] args) {
        Function<String, Integer> function = asFunction(String::length).andThen(n -> n * 2);
        System.out.println(function.apply("foo"));
    }

    public static <T1, T2> Function<T1, T2> asFunction(Function<T1, T2> function) {
        return function;     
    }
}

Answer 2

You can just save it into a variable:

Function<String, Integer> toLength = String::length;
Stream.of("ciao", "hola", "hello")
      .map(toLength.andThen(n -> n * 2));

Or you can use a cast, but it's less readable, IMO:

Stream.of("ciao", "hola", "hello")
      .map(((Function<String, Integer>) String::length).andThen(n -> n * 2));

Answer 3

You can use a cast

Stream.of("ciao", "hola", "hello")
        .map(((Function<String, Integer>) String::length)
                .andThen(n -> n * 2))
        .forEach(System.out::println);

prints

8
8
10

Answer 4

You may also use

Function.identity().andThen(String::length).andThen(n -> n * 2)

The problem is, String::length is not necessarily a Function; it can conform to many functional interfaces. It must be used in a context that provides target type, and the context could be - assignment, method invocation, casting.

If Function could provide a static method just for the sake of target typing, we could do

    Function.by(String::length).andThen(n->n*2)

static <T, R> Function<T, R> by(Function<T, R> f){ return f; }

---

For example, I use this technique in a functional interface

static <T> AsyncIterator<T> by(AsyncIterator<T> asyncIterator)

Syntax sugar to create an AsyncIterator from a lambda expression or a method reference.

This method simply returns the argument asyncIterator, which seems a little odd. Explanation:

Since AsyncIterator is a functional interface, an instance can be created by a lambda expression or a method reference, in 3 contexts:

 // Assignment Context
 AsyncIterator<ByteBuffer> asyncIter = source::read;
 asyncIter.forEach(...);

 // Casting Context
 ((AsyncIterator<ByteBuffer>)source::read)
     .forEach(...);

 // Invocation Context
 AsyncIterator.by(source::read)
     .forEach(...);

The 3rd option looks better than the other two, and that's the purpose of this method.

Answer 5

You could write:

Stream.of("ciao", "hola", "hello").map(String::length).map(n -> n * 2);

Answer 6

You should be able to achieve what you want inline by using casts:

Stream.of("ciao", "hola", "hello")
      .map(((Function<String, Integer>) String::length).andThen(n -> n * 2))

There are only 'type hints' for the compiler, so they don't actually 'cast' the object and don't have the overhead of an actual cast.

---

Alternatively, you can use a local variable for readability:

Function<String, Integer> fun = String::length

Stream.of("ciao", "hola", "hello")
      .map(fun.andThen(n -> n * 2));

---

A third way that may be more concise is with a utility method:

public static <T, X, U> Function<T, U> chain(Function<T, X> fun1, Function<X, U> fun2)
{
    return fun1.andThen(fun2);
}

Stream.of("ciao", "hola", "hello")
      .map(chain(String::length, n -> n * 2));

Please note that this is not tested, thus I don't know if type inference works correctly in this case.