CODEWITHSUNDEEP
javajava-streamjava-11
Stefan Haberl
Stefan Haberl

Reputation: 10539

Converting an Optional to a List

I want to collect an Optional into a List, so that I end up with a list with a single element, if the optional is present or an empty list if not.

They only way I came up with (in Java 11) is going through a Stream:

var maybe = Optional.of("Maybe");
var list = maybe.stream().collect(Collectors.toList());

I know, this should be rather efficient anyway, but I was wondering, if there is an easier way to convert the Optional into a List without using an intermediate Stream?

Upvotes: 5

Views: 5473

Answers (2)

fps
fps

Reputation: 34460

I think that the most idiomatic way would be to use Optional.map:

var maybe = Optional.of("Maybe");
var list = maybe.map(List::of).orElse(Collections.emptyList());

Or, if you don't want to create an empty list that might end up being not used at the end:

var list = maybe.map(List::of).orElseGet(Collections::emptyList);

Upvotes: 7

MC Emperor
MC Emperor

Reputation: 22977

Okay, I'll just post it as an answer.

List.of(maybe.get()). Of course this assumes that there actually is a value present within the optional. Otherwise: var list = maybe.isPresent() ? List.of(maybe.get()) : List.of().

Is it easier? Perhaps not. Is it without an intermediate stream? Yes.

Upvotes: 4

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