CODEWITHSUNDEEP
javahashmap
Braj
Braj

Reputation: 46841

HashMap having null as key

How HashMap internally differentiate between null and 0 as key.

As per this post the hash code for null key is 0, Is it correct? If yes then both should go in same bucket at index 0 and there should be one value in the HashMap but this is not how HashMap works.

Can somebody explain me? what is the hash code for null key in HashMap?

Sample code:

HashMap<Integer,String> map=new HashMap<Integer,String>();
map.put(null, "abc");
map.put(0, "xyz"); // will it override the value for null as key?

System.out.println(map.get(null));  // abc
System.out.println(map.get(0));     // xyz

Upvotes: 4

Views: 3327

Answers (4)

Stephen C
Stephen C

Reputation: 718758

If yes then both should go in same bucket at index 0 ...

Correct.

and there should be one value in the HashMap

Incorrect. There will be a separate entry for the two keys. This is guaranteed by the Map and HashMap specification; i.e. the javadocs say that there will be a separate entry for each distinct key. The implementation is required to meet the specification ... and it does.

Prior to Java 8, the way that HashMap handles collisions is to chain together all of the entries that hash to the same bucket. The HashMap.get(key) logic will iterate the chain for the relevant bucket, looking for an entry that matches the key. (In your example code, there would be separate entries in the chain for the null and the Integer(0) keys.)

In Java 8, it works the same to start with, but there is an optimization for the case where the keys all implement Comparable and the chains get too long. In this case, the long chains are converted into binary trees ... so that an O(N) chain scan turns into an O(logN) tree search.

Upvotes: 5

MirMasej
MirMasej

Reputation: 1622

This is possible because single Entry in the HashMap is not just key-value pair but also has a next field, so it's sort of simple linked list and adding of new entry looks as follows:

void createEntry(int hash, K key, V value, int bucketIndex) {
    Entry<K,V> e = table[bucketIndex];
    table[bucketIndex] = new Entry<>(hash, key, value, e);
    size++;
}

e local variable (the one that was already present) ends up as the next value in the Entry for given hash and bucketIndex.

Upvotes: 0

chenzhongpu
chenzhongpu

Reputation: 6871

You may have some mistakes with hash.

1, when the key is 0, the hash code is may not 0. Because the hash code is:

final int hash(Object k) {
    int h = hashSeed;
    if (0 != h && k instanceof String) {
        return sun.misc.Hashing.stringHash32((String) k);
    }
    h ^= k.hashCode();
    h ^= (h >>> 20) ^ (h >>> 12);
    return h ^ (h >>> 7) ^ (h >>> 4);
}

2, when your key is null, the JDK call a special method to do,

 if (key == null)
    return putForNullKey(value);

Yes, when key is null, the hash is 0. If you want know more, please see the JDK source code.

Upvotes: 0

harsh
harsh

Reputation: 7692

Both cases belongs to hashCode()==0 but equals differs hence two entries and both get return associated values.

Case with null:

public V put(null,"abc") {    
         if (key == null)
                    return putForNullKey(value);
}
private V putForNullKey(V value) {
        ..
        addEntry(0, null, value, 0); ==> 0th index with 0 hashCode
        return null;
    }

Case with 0:

public V put(K key, V value) {       
        int hash = hash(key.hashCode()); ==> 0
        int i = indexFor(hash, table.length); ==> 0
        for (Entry<K,V> e = table[i]; e != null; e = e.next) {
            Object k;
            if (e.hash == hash && ((k = e.key) == key || key.equals(k))) { ==> false
                ..
            }
        }

        modCount++;
        addEntry(hash, key, value, i); ==> addition to link list
        return null;
    }

Upvotes: 2

Related Questions