Why does HashSet say it does not contain this Object?

Question

I am trying to print "Contains", however, the HashSet is not detecting the Integer[] value. Does it have to do with it being Passed By Reference? How can I overcome this if I do not want to pass the actual Integer[] object into the method's arguments?

import java.util.*;

public class passByReference2{
    public static void method(HashSet<Integer[]> visited){
        Integer[] n = {1, 2};
        if (visited.contains(n)){
            System.out.println("Contains");
        }
    }

    public static void main(String[]args){
        HashSet<Integer[]> visited = new HashSet<Integer[]>();
        Integer[]v = {1, 2};
        visited.add(v);
        method(visited);
    }


}

*Nothing* in Java is passed by reference. It is pass-by-value only. — Andy Turner, May 16 '16 at 23:36

Andy Turner · Accepted Answer · 2016-05-16T23:37:36.533

3

It is because arrays don't implement hashCode in a way which compares elements - in fact, arrays don't override hashCode (or equals) at all, so the implementation from Object is used.

The hashCode and equals for an array are purely based on identity.

If you want to "overcome" it, use a type which calculates hashCode and equals based upon element values, e.g. ArrayList<Integer>.

edited May 16 '16 at 23:37

answered May 16 '16 at 23:34

Andy Turner

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They don't implement `equals`, either, so they inherit the default implementation that uses `==`. – user2357112 May 16 '16 at 23:36
@user2357112 true, updated. – Andy Turner May 16 '16 at 23:37

score 0 · Answer 2 · answered May 16 '16 at 23:38

In Java the default hashCode implementation is to use the address of the object.

This is typically implemented by converting the internal address of the object into an integer, but this implementation technique is not required by the JavaTM programming language.)

hashCode() javadocs

Why does HashSet say it does not contain this Object?

2 Answers2