容器-HashMap计算Hash值底层源码分析(十七)

容器-HashMap计算Hash值底层源码分析(十七)

1. 计算Hash值

   • 获得Key对象的hashcode

     首先调用key对象的hashcode()的方法，获得key的hashcode的值

   • 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，我们需要转化成[0,数组长度-1]的范围，我们要求转换后的hash值尽量均匀的分布在[0,数组长度-1]这个区间，减少“hash冲突”

2. 一种极端简单的算法是：

   • 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，这样就形成了一个非常长的链表。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"。

3. 一种简单和常用的算法是(相对取余法)

   • 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]这个区间，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)。(&：与运算，二进制的数中，两个为才为1，否则全为0).

4. 我们先去看put()的方法，只有添加元素，才会涉及到hash值的运算

   • map.put()方法用Ctrl+鼠标左键进入源代码，再用Ctrl+Alt选择put方法中的HashMap接口的实现类，进入源代码

     /**
          * 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);
         }
     

   • 然后我们看hash(key)方法

     /**
          * Computes key.hashCode() and spreads (XORs) higher bits of hash
          * to lower.  Because the table uses power-of-two masking, sets of
          * hashes that vary only in bits above the current mask will
          * always collide. (Among known examples are sets of Float keys
          * holding consecutive whole numbers in small tables.)  So we
          * apply a transform that spreads the impact of higher bits
          * downward. There is a tradeoff between speed, utility, and
          * quality of bit-spreading. Because many common sets of hashes
          * are already reasonably distributed (so don't benefit from
          * spreading), and because we use trees to handle large sets of
          * collisions in bins, we just XOR some shifted bits in the
          * cheapest possible way to reduce systematic lossage, as well as
          * to incorporate impact of the highest bits that would otherwise
          * never be used in index calculations because of table bounds.
          */
         static final int hash(Object key) {
             
             
             int h;
             return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
         }//主要看三目运算后面的那一个是什么，(h = key.hashCode()) ^ (h >>> 16)
     //先去取key的hashCode()的值赋给h，再用h做了一个右位移16的处理
     
     

   • (h = key.hashCode()) ^ (h >>> 16)//先去取key的hashCode()的值赋给h，再用h做了一个右位移16的处理，就是把32位的整数取了16位（右位移16位也就是从左到右去16位）

     ^(异或运算，相同为0，相异为1)

     假设hashCode()的值是：456789

     对应的二进制：0000 0000 0100 0101 0110 0111 1000 1001

     和右位移16位也就是从左到右去16位:0000 0000 0100 0101做^运算

• 然后返回到putVal()方法

    /**
       * 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)
              //(n - 1) & hash这个与运算就是为了计算哈希值的
              //n是我们之前分析出来的是16,16-1=15
              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;
      }
  
  

• 分析putVal()方法中计算哈希值的部分

 ```java
  if ((p = tab[i = (n - 1) & hash]) == null)
      //(n - 1) & hash这个与运算就是为了计算哈希值的
      //n是我们之前分析出来的是16,16-1=15
 ```

 15的二进制：0000 0000 0000 0000 0000 0000 0000 1111再去与hash做与运算

5. 总的一个分析：
   • 拿一个hashcode()的值先去转成二进制，搞成32位的整数，
   • 去前16位的数，在补齐32位，
   • 然后做异或运算，算出来是一个hash
   • hash和n-1的二进制做与运算
   • 得出hash值