How to add two Optional in java

Question

Optional<Long>totalLanding= ....(get it from somewhere);
Optional<Long>totalSharing = ...(get it from somewhere);

I want to do something like this not syntactically but logically

Optional<Long>total = totalLanding+totalSharing;

Such that if both are empty then total should be empty if one of them has the value then total should have that value is both of them have the value then they should get added and stored in total

Eran · Answer 1 · 2019-12-31T11:51:33.220

9

How about using Streams?

Optional<Long> total = Stream.of(totalLanding,totalSharing)
                             .filter(Optional::isPresent)
                             .map(Optional::get)
                             .reduce(Long::sum);

BTW, I'd use OptionalLong instead of Optional<Long>.

The solution would be similar:

OptionalLong total = Stream.of(totalLanding,totalSharing)
                           .filter(OptionalLong::isPresent)
                           .mapToLong(OptionalLong::getAsLong)
                           .reduce(Long::sum);

edited Dec 31 '19 at 11:51

answered Dec 31 '19 at 11:45

Eran

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You are forgetting that optional is a stream too. See my answer. – kan Dec 31 '19 at 12:30
1

@kan You mean that `Optional` has a `stream()` method. `Optional` is not a `Stream`. And it's not that I forgot, I'm just less familiar with methods added after Java 8. – Eran Dec 31 '19 at 12:32
Agree, I meant logically it's a stream too, alas Java 8 lacks it. Seems Java 8 API authors forgot it then ;) – kan Dec 31 '19 at 13:02

kan · Answer 2 · 2019-12-31T13:51:27.300

Java 9 or newer:

Optional<Long>total = Stream.concat(
        totalLanding.stream(),
        totalSharing.stream())
    .reduce(Long::sum)

Java 8 compatible variant:

Optional<Long>total = Stream.concat(
        totalLanding.map(Stream::of).orElseGet(Stream::empty),
        totalSharing.map(Stream::of).orElseGet(Stream::empty))
    .reduce(Long::sum)

Or better to extract the .map(Stream::of).orElseGet(Stream::empty) as a utility method and reuse. Or other variants here: How to convert an Optional<T> into a Stream<T>?

Costi Ciudatu · Answer 3 · 2019-12-31T18:51:02.400

0

This should work with Java 8:

Optional<Long> total = totalLanding.map(x -> x + totalSharing.orElse(0L))
        .map(Optional::of)   // wrap it twice for the next line to find an Optional inside
        .orElse(totalSharing);

edited Dec 31 '19 at 18:51

answered Dec 31 '19 at 12:39

Costi Ciudatu

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In Java 9 can be simplified to: `first.map(x -> x + second.orElse(0L)).or(() -> second);` – Alexander Terekhov Dec 31 '19 at 12:50
@AlexanderTerekhov That's true, thanks. I should've stated that this is a bit more verbose just so it compiles with Java 8. – Costi Ciudatu Dec 31 '19 at 12:54

score -1 · Answer 4 · answered Dec 31 '19 at 11:29

-1

How about

BigDecimal zero = BigDecimal.ZERO

Optional<Long> addition = Optional.of(totalLanding.orElse(zero).add(totalSharing.orElse(zero)));

answered Dec 31 '19 at 11:29

Sid

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Something like that was my idea too, but it wouldn't return an empty optional if neither was set. – findusl Dec 31 '19 at 11:36

malware · Answer 5 · 2019-12-31T12:04:15.087

-2

We can do like:

Optional<Long>result = Optional.of((totalLanding != null ? totalLanding.get() : 0L) + (totalSharing != null ? totalSharing.get() : 0L));

edited Dec 31 '19 at 12:04

answered Dec 31 '19 at 11:40

malware

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Have you actually tried that? This should lead to a NullPointerException if any of the Optionals is the result of `Optional.ofNullable(null)`. – Lothar Dec 31 '19 at 11:43
Also does not work for requirement _if both are empty then total should be empty_ – kan Dec 31 '19 at 13:48

How to add two Optional in java

5 Answers5