JAVA深化篇_21—— 面试必问之HashMap的底层源码分析

HashMap的底层源码分析

底层存储介绍

HashMap底层实现采用了哈希表，这是一种非常重要的数据结构。对于我们以后理解很多技术都非常有帮助。

数据结构中由数组和链表来实现对数据的存储，他们各有特点。

(1) 数组：占用空间连续。 寻址容易，查询速度快。但是，增加和删除效率非常低。

(2) 链表：占用空间不连续。 寻址困难，查询速度慢。但是，增加和删除效率非常高。

那么，我们能不能结合数组和链表的优点（即查询快，增删效率也高）呢？ 答案就是“哈希表”。 哈希表的本质就是“数组+链表”。

建议

对于本章中频繁出现的“底层实现”讲解，建议学有余力的童鞋将它搞通。刚入门的童鞋如果觉得有难度，可以暂时跳过。入门期间，掌握如何使用即可，底层原理是扎实内功，便于大家应对一些大型企业的笔试面试。

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HashMap中的成员变量

/**
 * 默认初始容量 - 必须是 2 的幂。
 */
static final int DEFAULT_INITIAL_CAPACITY = 1 << 4; // aka 16


/**
 * 最大容量，如果任一构造函数使用参数隐式指定了更高的值，则使用。必须是 2 的幂 <= 1<<30。
 */
static final int MAXIMUM_CAPACITY = 1 << 30;


/**
 * 构造函数中未指定时使用的荷载系数。
 */
static final float DEFAULT_LOAD_FACTOR = 0.75f;


/**
 * 使用树而不是库列表的箱计数阈值。将元素添加到至少具有这么多节点的箱时，箱将转换为树。
 * 该值必须大于 2，并且应至少为 8，以符合树木移除中关于在收缩时转换回普通箱的假设。
 * 
 */
static final int TREEIFY_THRESHOLD = 8;


/**
 * 在调整大小操作期间取消树化（拆分）箱的箱计数阈值。应小于TREEIFY_THRESHOLD，最多6个以啮合检测下去除。
 */
static final int UNTREEIFY_THRESHOLD = 6;


/**
 * 可以为其树化箱的最小表容量。
 *（否则，如果箱中的节点过多，则会调整表的大小。应至少为 4 * TREEIFY_THRESHOLD以避免调整大小和树化阈值之间的冲突。
 */
static final int MIN_TREEIFY_CAPACITY = 64;
/**
 * 此映射中包含的键值映射数
 */
transient int size;


/**
 * 表，在首次使用时初始化，并根据需要进行调整大小。分配时，长度始终是 2 的幂。
 *（我们还容忍某些操作中的长度为零，以允许当前不需要的引导机制。
 */
transient Node<K,V>[] table;

HashMap中存储元素的节点类型

Node类<单向链表>

static class Node<K,V> implements Map.Entry<K,V> {
  final int hash;
  final K key;
  V value;
  Node<K,V> next;


  Node(int hash, K key, V value, Node<K,V> next) {
    this.hash = hash;
    this.key = key;
    this.value = value;
    this.next = next;
   }


  public final K getKey()     { return key; }
  public final V getValue()    { return value; }
  public final String toString() { return key + "=" + value; }


  public final int hashCode() {
    return Objects.hashCode(key) ^ Objects.hashCode(value);
   }


  public final V setValue(V newValue) {
    V oldValue = value;
    value = newValue;
    return oldValue;
   }


  public final boolean equals(Object o) {
    if (o == this)
      return true;
    if (o instanceof Map.Entry) {
      Map.Entry<?,?> e = (Map.Entry<?,?>)o;
      if (Objects.equals(key, e.getKey()) &&
        Objects.equals(value, e.getValue()))
        return true;
     }
    return false;
   }
}

TreeNode类<红黑树>

/**
 * Entry for Tree bins. Extends LinkedHashMap.Entry (which in turn
 * extends Node) so can be used as extension of either regular or
 * linked node.
 */
static final class TreeNode<K,V> extends LinkedHashMap.Entry<K,V> {
  TreeNode<K,V> parent; // red-black tree links
  TreeNode<K,V> left;
  TreeNode<K,V> right;
  TreeNode<K,V> prev;  // needed to unlink next upon deletion
  boolean red;
  TreeNode(int hash, K key, V val, Node<K,V> next) {
    super(hash, key, val, next);
   }


  /**
   * Returns root of tree containing this node.
   */
  final TreeNode<K,V> root() {
    for (TreeNode<K,V> r = this, p;;) {
      if ((p = r.parent) == null)
        return r;
      r = p;
     }
   }

---

HashMap中的数组初始化

在JDK1.8的HashMap中对于数组的初始化采用的是延迟初始化方式。通过resize方法实现初始化处理。resize方法既实现数组初始化，也实现数组扩容处理。

/**
 * Initializes or doubles table size.  If null, allocates in
 * accord with initial capacity target held in field threshold.
 * Otherwise, because we are using power-of-two expansion, the
 * elements from each bin must either stay at same index, or move
 * with a power of two offset in the new table.
 *
 * @return the table
 */
final Node<K,V>[] resize() {
  Node<K,V>[] oldTab = table;
  int oldCap = (oldTab == null) ? 0 : oldTab.length;
  int oldThr = threshold;
  int newCap, newThr = 0;
  if (oldCap > 0) {
    if (oldCap >= MAXIMUM_CAPACITY) {
      threshold = Integer.MAX_VALUE;
      return oldTab;
     }
    else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
         oldCap >= DEFAULT_INITIAL_CAPACITY)
      newThr = oldThr << 1; // double threshold
   }
  else if (oldThr > 0) // initial capacity was placed in threshold
    newCap = oldThr;
  else {        // zero initial threshold signifies using defaults
    newCap = DEFAULT_INITIAL_CAPACITY;
    newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
   }
  if (newThr == 0) {
    float ft = (float)newCap * loadFactor;
    newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?
          (int)ft : Integer.MAX_VALUE);
   }
  threshold = newThr;
  @SuppressWarnings({"rawtypes","unchecked"})
    Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
  table = newTab;
  if (oldTab != null) {
    for (int j = 0; j < oldCap; ++j) {
      Node<K,V> e;
      if ((e = oldTab[j]) != null) {
        oldTab[j] = null;
        if (e.next == null)
          newTab[e.hash & (newCap - 1)] = e;
        else if (e instanceof TreeNode)
           ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
        else { // preserve order
          Node<K,V> loHead = null, loTail = null;
          Node<K,V> hiHead = null, hiTail = null;
          Node<K,V> next;
          do {
            next = e.next;
            if ((e.hash & oldCap) == 0) {
              if (loTail == null)
                loHead = e;
              else
                loTail.next = e;
              loTail = e;
             }
            else {
              if (hiTail == null)
                hiHead = e;
              else
                hiTail.next = e;
              hiTail = e;
             }
           } while ((e = next) != null);
          if (loTail != null) {
            loTail.next = null;
            newTab[j] = loHead;
           }
          if (hiTail != null) {
            hiTail.next = null;
            newTab[j + oldCap] = hiHead;
           }
         }
       }
     }
   }
  return newTab;
}

HashMap中计算Hash值

1. 获得key对象的hashcode

   首先调用key对象的hashcode()方法，获得key的hashcode值。

2. 根据hashcode计算出hash值(要求在[0, 数组长度-1]区间)hashcode是一个整数，我们需要将它转化成[0, 数组长度-1]的范围。我们要求转化后的hash值尽量均匀地分布在[0,数组长度-1]这个区间，减少“hash冲突”

   • 一种极端简单和低下的算法是：

     hash值 = hashcode/hashcode;

     也就是说，hash值总是1。意味着，键值对对象都会存储到数组索引1位置，这样就形成一个非常长的链表。相当于每存储一个对象都会发生“hash冲突”，HashMap也退化成了一个“链表”。

   • 一种简单和常用的算法是（相除取余算法）:

     hash值 = hashcode%数组长度;

     这种算法可以让hash值均匀的分布在[0,数组长度-1]的区间。但是，这种算法由于使用了“除法”，效率低下。JDK后来改进了算法。首先约定数组长度必须为2的整数幂，这样采用位运算即可实现取余的效果：hash值 = hashcode&(数组长度-1)。

/**
 * 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) {
  return putVal(hash(key), key, value, false, true);
}

static final int hash(Object key) {
  int h;
  return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}

/**
 * Implements Map.put and related methods
 *
 * @param hash hash for key
 * @param key the key
 * @param value the value to put
 * @param onlyIfAbsent if true, don't change existing value
 * @param evict if false, the table is in creation mode.
 * @return previous value, or null if none
 */
final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
        boolean evict) {
  Node<K,V>[] tab; Node<K,V> p; int n, i;
  if ((tab = table) == null || (n = tab.length) == 0)
    n = (tab = resize()).length;
  if ((p = tab[i = (n - 1) & hash]) == null)
    tab[i] = newNode(hash, key, value, null);
  else {
    Node<K,V> e; K k;
    if (p.hash == hash &&
       ((k = p.key) == key || (key != null && key.equals(k))))
      e = p;
    else if (p instanceof TreeNode)
      e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
    else {
      for (int binCount = 0; ; ++binCount) {
        if ((e = p.next) == null) {
          p.next = newNode(hash, key, value, null);
          if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
            treeifyBin(tab, hash);
          break;
         }
        if (e.hash == hash &&
           ((k = e.key) == key || (key != null && key.equals(k))))
          break;
        p = e;
       }
     }
    if (e != null) { // existing mapping for key
      V oldValue = e.value;
      if (!onlyIfAbsent || oldValue == null)
        e.value = value;
      afterNodeAccess(e);
      return oldValue;
     }
   }
  ++modCount;
  if (++size > threshold)
    resize();
  afterNodeInsertion(evict);
  return null;
}

HashMap中添加元素

/**
 * 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) {
  return putVal(hash(key), key, value, false, true);
}

/**
 * Implements Map.put and related methods
 *
 * @param hash hash for key
 * @param key the key
 * @param value the value to put
 * @param onlyIfAbsent if true, don't change existing value
 * @param evict if false, the table is in creation mode.
 * @return previous value, or null if none
 */
final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
        boolean evict) {
  Node<K,V>[] tab; Node<K,V> p; int n, i;
  if ((tab = table) == null || (n = tab.length) == 0)
    n = (tab = resize()).length;
  if ((p = tab[i = (n - 1) & hash]) == null)
    tab[i] = newNode(hash, key, value, null);
  else {
    Node<K,V> e; K k;
    if (p.hash == hash &&
       ((k = p.key) == key || (key != null && key.equals(k))))
      e = p;
    else if (p instanceof TreeNode)
      e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
    else {
      for (int binCount = 0; ; ++binCount) {
        if ((e = p.next) == null) {
          p.next = newNode(hash, key, value, null);
          if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
            treeifyBin(tab, hash);
          break;
         }
        if (e.hash == hash &&
           ((k = e.key) == key || (key != null && key.equals(k))))
          break;
        p = e;
       }
     }
    if (e != null) { // existing mapping for key
      V oldValue = e.value;
      if (!onlyIfAbsent || oldValue == null)
        e.value = value;
      afterNodeAccess(e);
      return oldValue;
     }
   }
  ++modCount;
  if (++size > threshold)
    resize();
  afterNodeInsertion(evict);
  return null;
}

HashMap中数组扩容

/**
 * Implements Map.put and related methods
 *
 * @param hash hash for key
 * @param key the key
 * @param value the value to put
 * @param onlyIfAbsent if true, don't change existing value
 * @param evict if false, the table is in creation mode.
 * @return previous value, or null if none
 */
final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
        boolean evict) {
  Node<K,V>[] tab; Node<K,V> p; int n, i;
  if ((tab = table) == null || (n = tab.length) == 0)
    n = (tab = resize()).length;
  if ((p = tab[i = (n - 1) & hash]) == null)
    tab[i] = newNode(hash, key, value, null);
  else {
    Node<K,V> e; K k;
    if (p.hash == hash &&
       ((k = p.key) == key || (key != null && key.equals(k))))
      e = p;
    else if (p instanceof TreeNode)
      e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
    else {
      for (int binCount = 0; ; ++binCount) {
        if ((e = p.next) == null) {
          p.next = newNode(hash, key, value, null);
          if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
            treeifyBin(tab, hash);
          break;
         }
        if (e.hash == hash &&
           ((k = e.key) == key || (key != null && key.equals(k))))
          break;
        p = e;
       }
     }
    if (e != null) { // existing mapping for key
      V oldValue = e.value;
      if (!onlyIfAbsent || oldValue == null)
        e.value = value;
      afterNodeAccess(e);
      return oldValue;
     }
   }
  ++modCount;
  if (++size > threshold)
    resize();
  afterNodeInsertion(evict);
  return null;
}

/**
 * Initializes or doubles table size.  If null, allocates in
 * accord with initial capacity target held in field threshold.
 * Otherwise, because we are using power-of-two expansion, the
 * elements from each bin must either stay at same index, or move
 * with a power of two offset in the new table.
 *
 * @return the table
 */
final Node<K,V>[] resize() {
  Node<K,V>[] oldTab = table;
  int oldCap = (oldTab == null) ? 0 : oldTab.length;
  int oldThr = threshold;
  int newCap, newThr = 0;
  if (oldCap > 0) {
    if (oldCap >= MAXIMUM_CAPACITY) {
      threshold = Integer.MAX_VALUE;
      return oldTab;
     }
    else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
         oldCap >= DEFAULT_INITIAL_CAPACITY)
      newThr = oldThr << 1; // double threshold
   }
  else if (oldThr > 0) // initial capacity was placed in threshold
    newCap = oldThr;
  else {        // zero initial threshold signifies using defaults
    newCap = DEFAULT_INITIAL_CAPACITY;
    newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
   }
  if (newThr == 0) {
    float ft = (float)newCap * loadFactor;
    newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?
          (int)ft : Integer.MAX_VALUE);
   }
  threshold = newThr;
  @SuppressWarnings({"rawtypes","unchecked"})
    Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
  table = newTab;
  if (oldTab != null) {
    for (int j = 0; j < oldCap; ++j) {
      Node<K,V> e;
      if ((e = oldTab[j]) != null) {
        oldTab[j] = null;
        if (e.next == null)
          newTab[e.hash & (newCap - 1)] = e;
        else if (e instanceof TreeNode)
           ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
        else { // preserve order
          Node<K,V> loHead = null, loTail = null;
          Node<K,V> hiHead = null, hiTail = null;
          Node<K,V> next;
          do {
            next = e.next;
            if ((e.hash & oldCap) == 0) {
              if (loTail == null)
                loHead = e;
              else
                loTail.next = e;
              loTail = e;
             }
            else {
              if (hiTail == null)
                hiHead = e;
              else
                hiTail.next = e;
              hiTail = e;
             }
           } while ((e = next) != null);
          if (loTail != null) {
            loTail.next = null;
            newTab[j] = loHead;
           }
          if (hiTail != null) {
            hiTail.next = null;
            newTab[j + oldCap] = hiHead;
           }
         }
       }
     }
   }
  return newTab;
}