Java hashmap方法.put（）

Question

The source code of java.util.HashMap#put(Object, Object) is

public V put(K key, V value) {

    if (key == null)
        return putForNullKey(value);
    int hash = hash(key.hashCode());
    int i = indexFor(hash, table.length);
    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))) {
            V oldValue = e.value;
            e.value = value;
            e.recordAccess(this);
            return oldValue;
        }
    }

    modCount++;
    addEntry(hash, key, value, i);
    return null;
}

I think that if the keys can get the same index in the table(example:key1 and key2 are all in the table[x]), the e.hash == hash will always return true. Is the code e.hash == hash neccessary for .put() method?

The only one explanation is that the different key.hashCode() after the operate of the hash() and indexFor() method, the result will be may the same, so the code will be necessary.

Is my answer right? Who can tell me, if e.hash == hash is necessary?

Answer 1

e.hash == hash is just an optimization; it's not strictly necessary. You're already comparing elements in the same hash bucket, and if two elements are .equals , then they match. The only point is minimizing expensive .equals checks on elements whose hashes are the same mod the hash table size.