Why IdentityHashMap.keySet().removeAll(keys) does not use identity? Is it a bug?

Question

public static void main(String[] args) {
    IdentityHashMap<Integer, Object> m1 = new IdentityHashMap<Integer, Object>();
    Integer ONE = 1;
    Integer TWO = 2;
    Integer OTHER_ONE = new Integer(1);
    Integer OTHER_TWO = new Integer(2);
    m1.put(ONE, new Object());
    m1.put(TWO, new Object());
    System.out.println(m1.keySet()); // [1, 2]

    m1.remove(OTHER_ONE); // Does not remove
    System.out.println(m1.keySet()); // [1, 2]

    m1.remove(ONE); // Does remove
    System.out.println(m1.keySet()); // [2]

    m1.keySet().removeAll(Arrays.asList(OTHER_TWO)); // ...
    System.out.println(m1.keySet()); // [] WHAT?
}

Reference here IdentityHashMap.keySet()

I found an answer in the source code (see below) but I don't know the ultimate reason. Is it a bug?

See http://stackoverflow.com/questions/2021374/boxed-primitives-and-equivalence/2021442. — Danstahr, Oct 30 '13 at 12:44
Danstahr i know that. I use new Integer() there intentionally to test identity behaviour of the map. I am asking about removeAll. Is the question not clear? Thank you. — aalku, Oct 30 '13 at 12:46

aalku · Accepted Answer · 2013-10-31T09:22:11.363

1

I found this source code from java 1.7.0_11

    public boolean removeAll(Collection<?> c) {
        boolean modified = false;
        for (Iterator<K> i = iterator(); i.hasNext(); ) {
            if (c.contains(i.next())) {
                i.remove();
                modified = true;
            }
        }
        return modified;
    }

It is using c.contains() so the behaviour depends on the collection class. I didn't understand that from the javadoc.

I found out it is a bug reported 6 years ago and still unresolved!! http://bugs.sun.com/bugdatabase/view_bug.do?bug_id=6588783

edited Oct 31 '13 at 09:22

answered Oct 30 '13 at 12:40

aalku

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There is a question in that answer, so maybe incorporate it into the original question. – siledh Oct 30 '13 at 13:02

Prabhakaran Ramaswamy · Answer 2 · 2013-10-30T13:05:04.017

-1

Try this one. It will illustrate what happened.

  System.out.println(OTHER_ONE == ONE);
  System.out.println(Arrays.asList(OTHER_ONE).contains( ONE));

OTHER_ONE is the object that will in the heap memory. ONE is not an object so it resides in the constant pool memory.

if you create Wrapper Objects with new keyword then only it will go to heap memory otherwise it will saved in the constant pool.

edited Oct 30 '13 at 13:05

answered Oct 30 '13 at 12:56

Prabhakaran Ramaswamy

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Because that's not what the question is about – siledh Oct 30 '13 at 13:18
My -1 is because it does not answer the question and ONE is an object too but you insist (don't edit) to say it is not what may confuse other programmers. – aalku Nov 02 '13 at 16:09

Why IdentityHashMap.keySet().removeAll(keys) does not use identity? Is it a bug?

2 Answers2