HashMap Internal Implementation Analysis in Java

Most of you will agree that HashMap is most favorite topic for discussion in interviews now-a-days. Now, I am continuing this discussion with you all.

I am assuming that if you are interested in internal working of HashMap, you already know basics of HashMap, so I'm skipping that part. But if you are new to concept, follow official java docs.

Single Statement Answer

If anybody asks me to describe "How HashMap works?", I simply answer: "On principle of Hashing". As simple as it is. Now before answering it, one must be very sure to know at least basics of Hashing. Right?

What is Hashing

Hashing in its simplest form, is a way to assigning a unique code for any variable/object after applying any formula/algorithm on its properties. A true Hashing function must follow this rule:

Hash function should return the same hash code each and every time, when function is applied on same or equal objects. In other words, two equal objects must produce same hash code consistently.

Note: All objects in java inherit a default implementation of hashCode() function defined in Object class. This function produce hash code by typically converting the internal address of the object into an integer, thus producing different hash codes for all different objects.

About Entry Class

A map by definition is : “An object that maps keys to values”. Very easy.. right?

So, there must be some mechanism in HashMap to store this key value pair. Answer is YES. HashMap has an inner class Entry, which looks like this:

static class Entry implements Map.Entry
        final K key;
        V value;
        Entry next;
        final int hash;
        ...//More code goes here

Surely Entry class has key and value mapping stored as attributes. Key has been marked as final and two more fields are there: next and hash. We will try to understand the need of these fields as we go forward.


What put() method actually does

Before going into put() method’s implementation, it is very important to learn that instances of Entry class are stored in an array. HashMap class defines this variable as:

 * The table, resized as necessary. Length MUST Always be a power of two.
transient Entry[] table;

Now look at code implementation of put() method:

 * Associates the specified value with the specified key in this map. If the
 * map previously contained a mapping for the key, the old value is
 * replaced.
 * @param key
 *            key with which the specified value is to be associated
 * @param value
 *            value to be associated with the specified key
 * @return the previous value associated with <tt>key</tt>, or <tt>null</tt>
 *         if there was no mapping for <tt>key</tt>. (A <tt>null</tt> return
 *         can also indicate that the map previously associated
 *         <tt>null</tt> with <tt>key</tt>.)
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 &amp;&amp; ((k = e.key) == key || key.equals(k))) {
            V oldValue = e.value;
            e.value = value;
            return oldValue;

    addEntry(hash, key, value, i);
    return null;

Lets note down the steps one by one:

  • First of all, key object is checked for null. If key is null, value is stored in table[0] position. Because hash code for null is always 0.
  • Then on next step, a hash value is calculated using key’s hash code by calling its hashCode() method. This hash value is used to calculate index in array for storing Entry object. JDK designers well assumed that there might be some poorly written hashCode() functions that can return very high or low hash code value. To solve this issue, they introduced another hash() function, and passed the object’s hash code to this hash() function to bring hash value in range of array index size.
  • Now indexFor(hash, table.length) function is called to calculate exact index position for storing the Entry object.
  • Here comes the main part. Now, as we know that two unequal objects can have same hash code value, how two different objects will be stored in same array location [called bucket].
    Answer is LinkedList. If you remember, Entry class had an attribute “next”. This attribute always points to next object in chain. This is exactly the behavior of LinkedList.

So, in case of collision, Entry objects are stored in LinkedList form. When an Entry object needs to be stored in particular index, HashMap checks whether there is already an entry? If there is no entry already present, Entry object is stored in this location.

If there is already an object sitting on calculated index, its next attribute is checked. If it is null, and current Entry object becomes next node in LinkedList. If next variable is not null, procedure is followed until next is evaluated as null.

What if we add the another value object with same key as entered before. Logically, it should replace the old value. How it is done? Well, after determining the index position of Entry object, while iterating over LinkedList on calculated index, HashMap calls equals method on key object for each Entry object. All these Entry objects in LinkedList will have similar hash code but equals() method will test for true equality. If key.equals(k) will be true then both keys are treated as same key object. This will cause the replacing of value object inside Entry object only.

In this way, HashMap ensure the uniqueness of keys.

How get() methods works internally

Now we have got the idea, how key-value pairs are stored in HashMap. Next big question is : what happens when an object is passed in get method of HashMap? How the value object is determined?

Answer we already should know that the way key uniqueness is determined in put() method , same logic is applied in get() method also. The moment HashMap identify exact match for the key object passed as argument, it simply returns the value object stored in current Entry object.

If no match is found, get() method returns null.

Let have a look at code:

 * Returns the value to which the specified key is mapped, or {@code null}
 * if this map contains no mapping for the key.
 * <p>
 * More formally, if this map contains a mapping from a key {@code k} to a
 * value {@code v} such that {@code (key==null ? k==null :
 * key.equals(k))}, then this method returns {@code v}; otherwise it returns
 * {@code null}. (There can be at most one such mapping.)
 * </p><p>
 * A return value of {@code null} does not <i>necessarily</i> indicate that
 * the map contains no mapping for the key; it's also possible that the map
 * explicitly maps the key to {@code null}. The {@link #containsKey
 * containsKey} operation may be used to distinguish these two cases.
 * @see #put(Object, Object)
public V get(Object key) {
    if (key == null)
        return getForNullKey();
    int hash = hash(key.hashCode());
    for (Entry<k , V> e = table[indexFor(hash, table.length)]; e != null; e = e.next) {
        Object k;
        if (e.hash == hash &amp;amp;&amp;amp; ((k = e.key) == key || key.equals(k)))
            return e.value;
    return null;

Above code is same as put() method till if (e.hash == hash && ((k = e.key) == key || key.equals(k))), after this simply value object is returned.

Load Factor of HashMap

 * The default initial capacity - MUST be a power of two.
static final int DEFAULT_INITIAL_CAPACITY = 16;

 * The load factor used when none specified in constructor.
static final float DEFAULT_LOAD_FACTOR = 0.75f;

It says default size of an array is 16 (always power of 2, we will understand soon why it is always power of 2 going further) and load factor means whenever the size of the HashMap reaches to 75% of its current size, i.e, 12, it will double its size by recomputing the hash codes of existing data structure elements.

Hence to avoid rehashing of the data structure as elements grow it is the best practice to explicitly give the size of the HashMap while creating it.

Do you foresee any problem with this resizing of hashmap in java? Since java is multi threaded it is very possible that more than one thread might be using same hashmap and then they both realize the need for re-sizing the hashmap at the same time which leads to race condition.

What is race condition with respect to hashmaps? When two or more threads see the need for resizing the same hashmap, they might end up adding the elements of old bucket to the new bucket simultaneously and hence might lead to infinite loops. FYI, in case of collision, i.e, when there are different keys with same same hashcode, internally we use single linked list to store the elements. And we store every new element at the head of the linked list to avoid tail traversing and hence at the time of resizing the entire sequence of objects in linked list gets reversed, during which there are chances of infinite loops.

HashTable and ConcurrentHashMap

This is another question which getting popular due to increasing popularity of ConcurrentHashMap. Since we knowHashtable is synchronized but ConcurrentHashMap provides better concurrency by only locking portion of map determined by concurrency level. ConcurrentHashMap is certainly introduced as HashTable and can be used in place of it but HashTable provide stronger thread-safety than ConcurrentHashMap.

ConcurrentHashMap do not allow null keys or null values while HashMap allows null keys.


  • http://howtodoinjava.com/2012/10/09/how-hashmap-works-in-java/
  • http://java.dzone.com/articles/hashmap-internal
  • http://javarevisited.blogspot.com/2011/02/how-hashmap-works-in-java.html
  • http://segmentfault.com/a/1190000003704860
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