Defining my own immutable array in Java

Question

I know that declaring an array final does not make it immutable.

I can however define a wrapper class that functions like an immutable array, e.g.

public class ImmutableIntArray {
    private int[] array;

    public ImmutableIntArray(int[] array) {
        this.array = (int []) array.clone();
    }

    public int get(int i) {
        return array[i];
    }

    public int length() {
        return array.length;
    }

    public static void main(String[] args) {
        ImmutableIntArray a = new ImmutableIntArray(new int[]{1, 2, 3});

        for (int i = 0; i < a.length(); ++i)
            System.out.println(a.get(i));
    }
}

This approach seems elegant to me, however, it seems like quite an obvious approach that I'm surprised I haven't seen anyone else apply it. Why shouldn't this be for example part of a standard library somewhere? Am I making any mistakes about this definition, so that my class is in fact mutable?

I believe the same approach would work for any Object which is immutable, that I could even define it using generics, i.e. an ImmutableArray<T> class.

you are simply removing some "privileges" to your array, you cannot add something to the array. In my opinion, you are just block some features to the user, but at the base there is still a mutable object. maybe I am wrong. — Gianmarco
– Gianmarco, Commented Sep 13, 2013 at 23:57
Igor, are Lists immutable? (is just a question because I don't know) — Gianmarco
– Gianmarco, Commented Sep 13, 2013 at 23:58
Your approach is throughly used, is extrange you didn't see it before. — Igor Rodriguez
– Igor Rodriguez, Commented Sep 13, 2013 at 23:59
Normal lists, such as ArrayList are mutable. But calling Collections.unmodifiableList(myListInstance) does the same as your class. It returns a List (it implements the interface) where mutators throw unsupportedOperationException. So is the same approach as yours. A good design. — Igor Rodriguez
– Igor Rodriguez, Commented Sep 14, 2013 at 0:02

nachokk · Accepted Answer · 2013-09-14 00:07:46Z

4

You are reinventing the wheel. You can have the same result using Collections#unmodifiableList(..).

Collections.unmodifiableList(Arrays.asList(yourArray));

answered Sep 14, 2013 at 0:07

nachokk

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