Functional composition of Optionals

Question

I have 2 Optionals (or Maybe objects) that I would like to combine so that I get the following results:

                ||       first operand 
    second      ++-------------+-------------
    operand     ||    empty    | optional(x)
    ============||=============|=============
    empty       ||    empty    | optional(x)
    ------------++-------------+-------------
    optional(y) || optional(y) |optional(x+y)

In other words, a non-empty Optional always replaces/overwrites an empty one, and two non-empty Optionals are combined according to some + function.

Initially, I assumed that the standard monadic flatMap method would do the trick, but (at least in Java) Optional.flatMap always returns an empty optional when the original Optional was already empty (and I'm not sure if any other implementation would comply with the Monad Laws).

Then, as both operands are wrapped in the same monadic type, I figured that this might be a good job for an Applicative Functor. I tried a couple different functional libraries, but I couldn't implement the desired behavior with any of the zip/ap methods that I tried.

What I'm trying to do seems to me a fairly common operation that one might do with Optionals, and I realize that I could just write my own operator with the desired behavior. Still, I am wondering if there is a standard function/method in functional programming to achieve this common operation?

Update: I removed the java tag, as I'm curious how other languages handle this situation

Answer 1

In a functional language, you'd do this with pattern matching, such as (Haskell):

combine :: Maybe t -> Maybe t -> (t -> t -> t) -> Maybe t
combine (Some x) (Some y) f = Some (f x y)
combine (Some x) _ _ = (Some x)
combine _ (Some y) _ = (Some y)
combine None None _ = None

There are other ways to write it, but you are basically pattern matching on the cases. Note that this still involves "unpacking" the optionals, but because its built into the language, it is less obvious.

Answer 2

It's not possible to combine optional objects without "unpacking" them.

I don't know the specifics of your case. For me, creating such a logic just in order to fuse the two optionals is an overkill.

But nevertheless, there's a possible solution with streams.

I assume that you're not going to pass optional objects as arguments (because such practice is discouraged). Therefore, there are two dummy methods returning Optional<T>.

Method combine() expects a BinaryOperator<T> as an argument and creates a stream by concatenating singleton-streams produced from each of the optional objects returned by getX() and getY().

The flavor of reduce(BinaryOperator) will produce an optional result.

public static <T> Optional<T> getX(Class<T> t) {
    return // something
}

public static <T> Optional<T> getY(Class<T> t) {
    return // something
}

public static <T> Optional<T> combine(BinaryOperator<T> combiner, 
                                      Class<T> t) {
    
    return Stream.concat(getX(t).stream(), getY(t).stream())
        .reduce(combiner);
}

If we generalize the problem to "how to combine N optional objects" then it can be solved like this:

@SafeVarargs
public static <T> Optional<T> combine(BinaryOperator<T> combiner,
                                      Supplier<Optional<T>>... suppliers) {
    
    return Arrays.stream(suppliers)
        .map(Supplier::get)           // fetching Optional<T>
        .filter(Optional::isPresent)  // filtering optionals that contain results to avoid NoSuchElementException while invoking `get()`
        .map(Optional::get)           // "unpacking" optionals
        .reduce(combiner);
}

Answer 3

In Haskell you can do this by wrapping any semigroup in a Maybe. Specifically, if you want to add numbers together:

Prelude> import Data.Semigroup
Prelude Data.Semigroup> Just (Sum 1) <> Just (Sum 2)
Just (Sum {getSum = 3})
Prelude Data.Semigroup> Nothing <> Just (Sum 2)
Just (Sum {getSum = 2})
Prelude Data.Semigroup> Just (Sum 1) <> Nothing
Just (Sum {getSum = 1})
Prelude Data.Semigroup> Nothing <> Nothing
Nothing

The above linked article contains more explanations, and also some C# examples.

Answer 4

Here's one way:

a.map(x -> b.map(y -> x + y).orElse(x)).or(() -> b)

Ideone Demo

Answer 5

OptionalInt x = ...
OptionalInt y = ...

OptionalInt sum = IntStream.concat(x.stream(), y.stream())
    .reduce(OptionalInt.empty(),
        (opt, z) -> OptionalInt.of(z + opt.orElse(0)));

Since java 9 you can turn an Optional into a Stream. With concat you get a Stream of 0, 1 or 2 elements.

Reduce it to an empty when 0 elements,and for more add it to the previous OptionalInt, defaulting to 0.

Not very straight (.sum()) because of the need for an empty().

Answer 6

You can implement your function in Java by combining flatMap and map:

optA.flatMap(a -> optB.map(b -> a + b));

More general example:

public static void main(String[] args) {
    test(Optional.empty(), Optional.empty());
    test(Optional.of(3), Optional.empty());
    test(Optional.empty(), Optional.of(4));
    test(Optional.of(3), Optional.of(4));
}

static void test(Optional<Integer> optX, Optional<Integer> optY) {
    final Optional<Integer> optSum = apply(Integer::sum, optX, optY);
    System.out.println(optX + " + " + optY + " = " + optSum);
}

static <A, B, C> Optional<C> apply(BiFunction<A, B, C> fAB, Optional<A> optA, Optional<B> optB) {
    return optA.flatMap(a -> optB.map(b -> fAB.apply(a, b)));
}

Since flatMap and map are standard functions for Optional/Maybe (and monad types generally), this approach should work in any other language (though most FP languages will have a more concise solution). E.g. in Haskell:

combine ma mb = do a <- ma ; b <- mb ;  return (a + b)

Answer 7

In F#, i would call this logic reduce.

Reason:

• The function must be of type 'a -> 'a -> 'a as it only can combine thinks of equal type.

• Like other reduce operations, like on list, you always need at least one value, otherwise it fails.

With a option and two of them, you just need to cover four cases. In F# it will be written this way.

(* Signature: ('a -> 'a -> 'a) -> option<'a> -> option<'a> -> option<'a> *)
let reduce fn x y =
    match x,y with
    | Some x, Some y -> Some (fn x y)
    | Some x, None   -> Some x
    | None  , Some y -> Some y
    | None  , None   -> None

printfn "%A" (reduce (+) (Some 3) (Some 7)) // Some 10
printfn "%A" (reduce (+) (None)   (Some 7)) // Some 7
printfn "%A" (reduce (+) (Some 3) (None))   // Some 3
printfn "%A" (reduce (+) (None)   (None))   // None

In another lets say Pseudo-like C# language, it would look like.

Option<A> Reduce(Action<A,A,A> fn, Option<A> x, Option<A> y) {
    if ( x.isSome ) {
        if ( y.isSome ) {
            return Option.Some(fn(x.Value, y.Value));
        }
        else {
            return x;
        }
    }
    else {
        if ( y.isSome ) {
            return y;
        }
        else {
            return Option.None;
        }
    }
}