How to initialize List in one line?

Question

How to initialize a list of short elements in one line?

Something like this:

List<Short> numbers = new ArrayList<Short>(List.of(1, 2, 3, 4, 5));

But I am getting:

"Cannot resolve constructor 'ArrayList(java.util.List<E>)'"

score 3 · Answer 1 · answered Jul 18 '20 at 09:45

3

The type ArrayList is declared as:

public class ArrayList<E> ... {
    public ArrayList(Collection<? extends E> c) { ... }
}

That means in your statement

List<Short> numbers = new ArrayList<>( ... );

the argument that you pass into the constructor must be of type Collection<? extends Short>. But you are passing a

List.of(1, 2, 3, 4, 5)

which is of type List<Integer>. Note that integral literals are int by default. Unfortunately, there is no short way to make them short. But you can always cast them:

List<Short> numbers = new ArrayList<>(List.of((short) 1, (short) 2, (short) 3, (short) 4, (short) 5));

This compiles fine.

Side note: Why do you need to create another list object (new ArrayList(...)) while already having one (List.of(...))?

answered Jul 18 '20 at 09:45

Seelenvirtuose

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`List.of` creates an immutable list. Wrapping it into a new `ArrayList` makes it mutable again – QBrute Jul 18 '20 at 09:59
@Seelenvirtuose Thank you for the answer. I am creating another list object, because I want to explicitly tell JVM that I want my list to be an Array List, due to accessing elements. – Методи Владимиров Jul 18 '20 at 10:04
Hmmm ... But why? – Seelenvirtuose Jul 18 '20 at 10:08
@Seelenvirtuose Because if I leave it only as `List.of` my accessing an element operation is slower rather than using `ArrayList`. – Методи Владимиров Jul 18 '20 at 10:13
No, that is not true! – Seelenvirtuose Jul 18 '20 at 10:20
@Seelenvirtuose but why? – Методи Владимиров Jul 18 '20 at 10:36
The question is, why do you think, that one is _slower_ than the other! – Seelenvirtuose Jul 18 '20 at 14:58

score 0 · Accepted Answer · answered Jul 18 '20 at 10:03

0

Use Integer.shortValue() method: Like below

  List<Short> numbers = Stream.of(1, 2, 3, 4, 5).map(value -> value.shortValue()).collect(Collectors.toList());

answered Jul 18 '20 at 10:03

Pandey Amit

657
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How to initialize List in one line?

2 Answers2