How to map an OptionalLong to an Optional?

Question

I have an instance of OptionalLong. But one of my libraries requires an Optional<Long> as a parameter.

How can I convert my OptionalLong into an Optional<Long>?

I was dreaming about something like this:

OptionalLong secondScreenHeight = OptionalLong.of(32l); // or: OptionalLong.empty()
api.setHeight(secondScreenHeight.mapToRegularOptional()); // .mapToUsualOptional does not exist

There are many variations, but most of them leverage `OptionalLong.isPresent()` to distinguish between the empty and non-empty cases. — John Bollinger, Aug 13 '18 at 13:18
the question is why doesn't `OptionalLong` have no `map/filter` etc — Eugene, Aug 13 '18 at 13:24
`api.setHeight(…)` should not expect an `Optional` as argument type. Then you will not have such problems. — Holger, Aug 13 '18 at 13:39
@Holger In some cases it cannot easily be avoided (i.e. if `Long` is the value of a type argument of a generic method) — slartidan, Aug 13 '18 at 13:42
Just don’t use `Optional` as parameter at all, whether generic or not. It's horrible to read. What does it mean to call `setHeight` with an empty optional argument? Do nothing? Then, don’t call it at all. `secondScreenHeight.ifPresent(api::setHeight)` makes much more sense. — Holger, Aug 13 '18 at 13:49
To add to what Holger already pointed out, this is what IntelliJ would suggest you as well... *...Optional was designed to provide a limited mechanism for library method return types where there needed to be a clear way to represent "no result". Using a field with type java.util.Optional is also problematic if the class needs to be Serializable, [which java.util.Optional is not.](https://stackoverflow.com/questions/24547673/why-java-util-optional-is-not-serializable-how-to-serialize-the-object-with-suc)* — Naman, Aug 13 '18 at 16:37
Why a bounty is opened here? The question received good answers — Mạnh Quyết Nguyễn, Aug 19 '18 at 11:53

score 14 · Answer 1 · answered Aug 13 '18 at 13:20

14

You could do this:

final OptionalLong optionalLong = OptionalLong.of(5);

final Optional<Long> optional = Optional.of(optionalLong)
            .filter(OptionalLong::isPresent)
            .map(OptionalLong::getAsLong);

answered Aug 13 '18 at 13:20

marstran

26,413
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1

What I like about this solution is, that the initial variable is only used once. This makes it possible to create "long chains" in my code - I do not necessarily store the value in a variable. – slartidan Aug 13 '18 at 13:35
1

I like that it does not use old `if else` syntax, but only shiny new java-8 syntax! – slartidan Aug 13 '18 at 13:36
@slartidan well, still `if/else` inside filter and map though – Eugene Aug 13 '18 at 13:37

Orest Savchak · Accepted Answer · 2018-08-13T13:20:54.053

10

I don't know simpler solutions but this will do what you need.

OptionalLong secondScreenHeight = OptionalLong.of(32l);
Optional<Long> optional = secondScreenHeight.isPresent() 
    ? Optional.of(secondSceenHeight.getAsLong()) 
    : Optional.empty();
api.setHeight(optional);

edited Aug 13 '18 at 13:20

answered Aug 13 '18 at 13:15

Orest Savchak

4,529
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I like about this solution, that it is readable and that the intent to remap the optional is clearly visible. – slartidan Aug 13 '18 at 14:17

df778899 · Answer 3 · 2018-08-19T19:01:25.797

5

One more possibility, though only from JDK 9 is via the new OptionalLong.stream() method, which returns a LongStream. This can then be boxed to a Stream<Long>:

OptionalLong optionalLong = OptionalLong.of(32);
Optional<Long> optional = optionalLong.stream().boxed().findFirst();

With JDK 8 something similar can be done, by stepping out to the Streams utility class in Guava:

Optional<Long> optional = Streams.stream(optionalLong).boxed().findFirst();

edited Aug 19 '18 at 19:01

answered Aug 19 '18 at 18:50

df778899

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1
24
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5

Also, `findAny` is 2 characters shorter. – Radiodef Aug 19 '18 at 22:36
I like about it, that it is short and does not rely on `isPresent`. However the intention is not as clearly visible as in some of the other answers. Thanks for sharing this solution! – slartidan Aug 20 '18 at 09:58

score 1 · Answer 4 · answered Aug 13 '18 at 13:16

1

This should work.

Optional<Long> returnValue = Optional.empty();
if(secondScreenHeight.isPresent()) {
      returnValue = Optional.of(secondScreenHeight.getAsLong());
}

answered Aug 13 '18 at 13:16

pvpkiran

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How to map an OptionalLong to an Optional?

4 Answers4

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