Why do I have to return Unit.INSTANCE when implementing in Java a Kotlin function that returns a Unit?

Question

If I have a Kotlin function

fun f(cb: (Int) -> Unit)

and I want to call f from Java, I have to do it like:

f(i -> {
     dosomething();
     return Unit.INSTANCE;
});

which looks very ugly. Why can't I just write it like f(i -> dosomething());, since Unit in Kotlin is equivalent to void in Java?

Answer 1

Unit in Kotlin is mostly equivalent to void in Java, however only when the rules of the JVM allow it.

Functional types in Kotlin are represented by interfaces like:

public interface Function1<in P1, out R> : Function<R> {
    /** Invokes the function with the specified argument. */
    public operator fun invoke(p1: P1): R
}

When you declare (Int) -> Unit, from Java's point of view this is equivalent to Function<Integer, Unit>. That's why you have to return a value. To work around this problem, in Java there are two separate interfaces Consumer<T> and Function<T, R> for when you don't have/have a return value.

The Kotlin designers decided to forgo the duplication of functional interfaces and instead rely on compiler "magic". If you declare a lambda in Kotlin, you don't have to return a value because the compiler will insert one for you.

To make your life a little bit easier, you can write a helper method that wraps a Consumer<T> in a Function1<T, Unit>:

public class FunctionalUtils {
    public static <T> Function1<T, Unit> fromConsumer(Consumer<T> callable) {
        return t -> {
            callable.accept(t);
            return Unit.INSTANCE;
        };
    }
}

Usage:

f(fromConsumer(integer -> doSomething()));

---

Fun fact: The special handling of Unit by the Kotlin compiler is the reason you can write code like:

fun foo() {
    return Unit
}

or

fun bar() = println("Hello World")

Both methods have return type void in the generated bytecode but the compiler is smart enough to figure that out and allow you to use return statements/expressions anyway.