why ToDoubleBiFunction doesn't extend BiFunction

Question

I wonder why ToDoubleBiFunction doesn't extend BiFunction. Just how I did:

import java.util.function.BiFunction;

public class Test {
  public static void main(String[] args) {
    MyToDoubleBiFunction<Integer, Integer> f = (a, b) -> a.doubleValue() + b.doubleValue();
    System.out.println(f.apply(1, 2));
  }
}

@FunctionalInterface
interface MyToDoubleBiFunction<T, U> extends BiFunction<T, U, Double> {
}

This code works and prints 3.0.

Also what is the point of using wordy applyAsDouble method instead of well-known apply?

The result type of `MyToDoubleBiFunction.apply` is `Double` whereas it is `double` for `ToDoubleBiFunction.applyAsDouble`. — LuCio, Sep 27 '18 at 20:24

score 4 · Accepted Answer · answered Sep 27 '18 at 20:17

4

The point of specific versions of BiFunction such as ToDoubleBiFunction is to avoid auto(un)boxing - in this case, autoboxing / -unboxing a primitive double value to / from a java.lang.Double object.

If ToDoubleBiFunction would extend BiFunction<T, U, Double> then you'd have auto(un)boxing.

Note that you cannot use primitive types as type arguments in Java, so BiFunction<T, U, double> (note that double is the primitive type here) is not possible.

answered Sep 27 '18 at 20:17

Jesper

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Nice, concrete, and efficient answer. +1 from me. What I'm wondering (without diving to the source code) how WE can write the `Function` without auto(un)boxing with three, four, or more parameters? e.g. `ToIntThreeFunction`... – zlakad Sep 27 '18 at 20:29
@zlakad there's nothing magical: you just define an interface with a single method, returning an `int`: `interface ToIntThreeFunction { int apply3(A a, B b, C c); }`. – Andy Turner Sep 27 '18 at 20:31
@AndyTurner, ah, well, I'll try to that. TY. – zlakad Sep 27 '18 at 20:32

why ToDoubleBiFunction doesn't extend BiFunction

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