Three possible java8 Optional values - how do I cleanly return whichever is present?

Question

I have three java8 Optionals, and want to return whichever one is actually present in a preferred order. It seems like there should be an easy way to chain them like so:

return optionalA.orElseIfPresent(optionalB).orElseIfPresent(optionalC);

if all three are empty, then an Optional.empty() ought to be returned.

the existing orElse and orElseGet are not really up to the task - they must return an actual value, so it's not possible for the remaining fallbacks to be Optionals themselves.

In the worst case I can have a long list of ifPresent() checks, but there just seems like there's a better way to go about it?

Answer 1

return Stream.of(optional1, optional2, optional3)
             .filter(Optional::isPresent)
             .map(Optional::get)
             .findFirst();

Answer 2

I like JB Nizet's answer (upvoted as well). Alternatively, using only Optional (for whatever reason):

Optional.ofNullable(
    optionalA.orElse(
        optionalB.orElse(
            optionalC.orElse(null))
        )
    );

which falls into indentation/parenthesis madness and I personally do not like.