How does java.util.function.Function.compose() work?

Question

The functional interface Function in Java 1.8 implements compose() as shown:

default <V> Function<V, R> compose(Function<? super V, ? extends T> before) {
    Objects.requireNonNull(before);
    return (V v) -> apply(before.apply(v));
}

My understanding of lambda expressions (possibly incorrect) is that the above return statement is syntactic sugar for:

    return new Function<V, R>() {
        R apply(V v) {
            return apply(before.apply(v));
        }};

This statement would be illegal because, as an interface, Function<V, R> should not be instantiable. So how does the above method work?

Boann · Accepted Answer · 2015-03-15 19:33:52Z

2

No, it's not syntactic sugar for an anonymous class.

And even if it was, the code you posted doesn't instantiate an interface. It instantiates an anonymous class which implements that interface.

Read the Java tutorial about anonymous inner classes.

edited Mar 15, 2015 at 19:33

Boann

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answered Mar 13, 2015 at 21:28

JB Nizet

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1 Comment

Ellen Spertus Over a year ago

Ah, right. We're providing the one non-default function. Thanks.

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