One liner for comparing by property and returning a stream?

Is it possible to sort and return a stream in one line as in the following?

//@Getter...
public List<Actor> actors;

public Stream<Actor> actors() {
    //can these two lines be combined somehow?
    actors.sort(Comparator.comparing(Actor::getName));
    return actors.stream();
}

Note: After posting this question, I gather this would lead to a design that is not ideal and perhaps error-prone, as it would mutate state outside of the method. Thanks to @knittl for pointing this out.

Streams has a sorted method

actors.stream().sorted(Comparator.comparing(Actor::getName))

Note: This will not sort the List<Actors>. And will only sort the returned stream.

you can do it like this:

return actors.stream().sorted(Comparator.comparing(Actor::getName));

The sort Method of a List will return void, so there is no chance to concat.

Greetings

If you really, really must:

public Stream<Actor> actors() {
    return (actors = actors.stream()
        .sorted(Comparator.comparing(Actor::getName))
        .toList()).stream();
}

Because it is possible, doesn't mean you should do this.

But indeed, knittl is right. You shouldn't mutate actors with this getter method. I'd be really surprised if I were to use this method and suddenly actors was sorted.

There are some options which are better alternatives to your current design:

• Sort the resulting stream, but don't mutate actors:

  return actors.stream()
      .sorted(Comparator.comparing(Actor::getName));
  

• Sort actors on initialization, for example:

  List<Actor> actors;
  
  YourConstructor(List<Actor> actors) {
      this.actors = actors.stream()
          .sorted(Comparator.comparing(Actor::getName))
          .toList(); // Or whatever implementation of List is desired
  }