'Java - Add one element to an immutable list
I need an immutable list where I can get derive a second immutable list preserving all elements of the previous list plus an additional element in Java (without additional libraries).
Note: This question is similar to What is an efficient and elegant way to add a single element to an immutable set? but I need a list and don't have Guava.
What I have tried so far:
var list = List.of(someArrayOfInitialElements);
var newList = Stream.concat(list.stream(), Stream.of(elementToAppend))
.collect(CollectorsCollectors.toUnmodifiableList());
That would work but creating a stream and copying elements one by one seems inefficient to me. You could basically bulk copy memory given that List.of() stores data in a field-based or array-based data structure.
Is there a more efficient solution than using streams? A better data structure in the Java standard library that I am missing?
Solution 1:[1]
The Answer by Frisch is correct, and should be accepted. One further noteā¦
Calling Collections.unmodifiableList produces a collection that is a view onto the original mutable list. So a modification to the original list will "bleed through" to the not-so-immutable second list.
This issue does not apply to the correct code shown in that Answer, because the new ArrayList object deliberately goes out-of-scope. Therefore that new list cannot be accessed for modification. But in other coding scenarios, this issue could be a concern.
List.copyOf
If you want an independent and truly immutable second list, use List.copyOf in Java 10+. This returns an unmodifiable list.
return List.copyOf( bl ) ;
Solution 2:[2]
Both answers are great, I would create a bit more generic solution:
private static <T> List<T> append(final List<T> al, final T... ts) {
final List<T> bl = new ArrayList<>(al);
for (final T t : ts) {
bl.add(t);
}
return List.copyOf(bl);
}
It can be used exactly like previous answer:
List<String> al = append(new ArrayList<>(), "1");
List<String> bl = append(al, "2");
System.out.println(bl);
But also slightly more efficient:
List<String> bl = append(new ArrayList<>(), "1", "2");
System.out.println(bl);
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Basil Bourque |
| Solution 2 | Rob Audenaerde |
