Fork me on GitHub

Thinking in java 基础之HashMap

java

一个简单的put引发的惨案

版本 jdk1.8
Map hashMap = new HashMap<String,String>();
hashMap.put("abc","1");
一个简单的put

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
/**
 * 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);
}

怎么进行hash

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
/**
    * 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);
   }

key可以为null ，返回0，直接落在第一个坑位上。

解释了key可以为null
Object.hashCode();

1
public native int hashCode();

以String举例
String.hashCode();

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
/**
    * Returns a hash code for this string. The hash code for a
    * {@code String} object is computed as
    * <blockquote><pre>
    * s[0]*31^(n-1) + s[1]*31^(n-2) + ... + s[n-1]
    * </pre></blockquote>
    * using {@code int} arithmetic, where {@code s[i]} is the
    * <i>i</i>th character of the string, {@code n} is the length of
    * the string, and {@code ^} indicates exponentiation.
    * (The hash value of the empty string is zero.)
    *
    * @return  a hash code value for this object.
    */
   public int hashCode() {
       int h = hash;
       if (h == 0 && value.length > 0) {
           char val[] = value;

           for (int i = 0; i < value.length; i++) {
               h = 31 * h + val[i];
           }
           hash = h;
       }
       return h;
   }

举例

1
2
System.out.println("abc".hashCode());//96354
System.out.println(97*31*31 + 98*31+99);//96354

s[0]31^(n-1) + s[1]31^(n-2) + … + s[n-1]

String的hash算法,那么HashMap中的hash(),
(key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
“abc”进行转hashCode()之后就是96354，然后变成二进制

1
2
3
4
5
6
7
8
h>>>16   h^(h>>>16)
那么就是（前面补0）
96354       0000 0000 0000 0001 0111 1000 0110 0010  低16位
h>>>16      0000 0000 0000 0000 0000 0000 0000 0001  高16位
^--------------------------------------------------
h^(h>>>16)  0000 0000 0000 0001 0111 1000 0110 0011  

最终得到96355

高16位与低16位

结果验证
hash运行时

然后下一步

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
/**
     * 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;
    }

没有数组

看内部的如果table是空的创建

1
2
if ((tab = table) == null || (n = tab.length) == 0)
           n = (tab = resize()).length;

resize()把table创建好了，同时把tab.length赋给n

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
/**
   * 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 这里是初始化的时候，放容量Cap=16和最多放几个开始扩容负载因子Thr = 0.75*16 = 12，
          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"})
      //创建一个newCap容量大小的数组，然后赋给table
          Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
      table = newTab;
      //以下内容为把原来的Node放到新的Node中，先放一下，后面在扩容判定位置的时候再研究
      if (oldTab != null) {
          ...
          ...
          ...
      }
      return newTab;
  }

先看新建的，不看扩容的。知道table是个newCap大小的数组

hash运行时

数组对应的位置为空

1
2
if ((p = tab[i = (n - 1) & hash]) == null)
           tab[i] = newNode(hash, key, value, null);

有了数组，怎么计算放在哪一个数组上，(n - 1) & hash],这个位置上如果为空，那么tab的i上就放当前的Node<K,V>;当前的n就是tab.length = 16
算一下

1
2
3
4
5
6
7
8
刚才abc的hash值是96355
h^(h>>>16)  0000 0000 0000 0001 0111 1000 0110 0011   96355
16-1        0000 0000 0000 0000 0000 0000 0000 1111   15
&-------------------------------------------------- 保证所有的值都落在tab之内
(n-1)hash   0000 0000 0000 0000 0000 0000 0000 0011   3

然后看tab[3]上是否为null,为空直接赋值在这里
然后tab[3] =  Node<>(hash, key, value, next);//Node<>(96355，"abc","1",null)

hash运行时

数组对应的位置不为空，跟链表，超过TREEIFY_THRESHOLD时，当前hash对应的位置转为treeifyBin

1
2
3
4
Node<K,V> e; K k;
if (p.hash == hash &&
    ((k = p.key) == key || (key != null && key.equals(k))))
    e = p;

判断key是不是一样的，如果是一样的，直接替换前面的Node，

解释了 为什么key可以为null，但key一样的话不能重复

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
else {
    //没有条件的一直循环，在内部break，
    for (int binCount = 0; ; ++binCount) {
        if ((e = p.next) == null) {
            p.next = newNode(hash, key, value, null);
            //当前容量大于8-1 = 7 的时候进行转为红黑树
            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;
    }
}

当前hash对应的位置转TreeNode

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
/**
     * Replaces all linked nodes in bin at index for given hash unless
     * table is too small, in which case resizes instead.
     */
    final void treeifyBin(Node<K,V>[] tab, int hash) {
        int n, index; Node<K,V> e;
        //如果tab数组小于MIN_TREEIFY_CAPACITY=64重新调整大小
        if (tab == null || (n = tab.length) < MIN_TREEIFY_CAPACITY)
            resize();
        else if ((e = tab[index = (n - 1) & hash]) != null) {
            TreeNode<K,V> hd = null, tl = null;
            do {
                TreeNode<K,V> p = replacementTreeNode(e, null);
                if (tl == null)
                    hd = p;
                else {
                    p.prev = tl;
                    tl.next = p;
                }
                tl = p;
            } while ((e = e.next) != null);
            if ((tab[index] = hd) != null)
                hd.treeify(tab);
        }
    }

数组对应的位置不为空，跟红黑树

直接调用TreeNode的putTreeVal()

1
2
else if (p instanceof TreeNode)
    e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);

里面对应的旋转啊，啥子的，一堆。

超过最大值，要进行扩容

当大小超过threshold=DEFAULT_LOAD_FACTOR DEFAULT_INITIAL_CAPACITY= 0.7516=12重新resize()

1
2
if (++size > threshold)
        resize();

resize(),扩容的时候MAXIMUM_CAPACITY=1<<30

1
2
3
4
5
6
7
8
9
if (oldCap >= MAXIMUM_CAPACITY) {
    threshold = Integer.MAX_VALUE;//最大值
    return oldTab;
}
//不到最大值，新的容量newCap为原来的左移一位 oldThr<<1, 两倍的容量
else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
            oldCap >= DEFAULT_INITIAL_CAPACITY)
    newThr = oldThr << 1; // double threshold
    //newThr 为12<<1 为 24

扩容怎么判定原来的位置，怎么移动

上面扩容省略的部分

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
for (int j = 0; j < oldCap; ++j) {
        Node<K,V> e;
        if ((e = oldTab[j]) != null) {
            oldTab[j] = null;
            //没有next 说明只有一个Node直接在table数组中直接移动
            if (e.next == null)
                newTab[e.hash & (newCap - 1)] = e;
            //是树的情况
            else if (e instanceof TreeNode)
                ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
            else { // preserve order 顺序的模式链表
                Node<K,V> loHead = null, loTail = null;
                Node<K,V> hiHead = null, hiTail = null;
                Node<K,V> next;
                do {
                    next = e.next;
                    //hash与oldCap容量如果等于0 则还在原来的坑位，
                    if ((e.hash & oldCap) == 0) {
                        if (loTail == null)
                            loHead = e;
                        else
                            loTail.next = e;
                        loTail = e;
                    }
                    //在新的坑位
                    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;
                }
            }
        }
    }

((TreeNode<K,V>)e).split(this, newTab, j, oldCap);中有一段，当小于

1
2
3
4
5
6
7
if (lc <= UNTREEIFY_THRESHOLD)
    tab[index] = loHead.untreeify(map);
else {
    tab[index] = loHead;
    if (hiHead != null) // (else is already treeified)
        loHead.treeify(tab);
}

已经把原来各个节点Node移动到新的数组中，当小于UNTREEIFY_THRESHOLD=6时，转为链表

链表的情况(e.hash & oldCap) == 0

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
刚才abc的hash值是96355
h^(h>>>16)  0000 0000 0000 0001 0111 1000 0110 0011   96355
16=oldCap   0000 0000 0000 0000 0000 0000 0001 0000   16
&-------------------------------------------------- 保证所有的值都落在tab之内
hash&oldCap 0000 0000 0000 0000 0000 0000 0000 0000   0

等于0 还在原来的位置


h^(h>>>16)  0000 0000 0000 0001 0111 1000 0110 0011
16扩32      0000 0000 0000 0000 0000 0000 0001 1111
&-------------------------------------------------- 保证所有的值都落在tab之内
hash&oldCap 0000 0000 0000 0000 0000 0000 0000 0011   3

如果不是0的话则在 16+3的位置

线程不安全

ConCurrentHashMap是怎么加锁的

ConcurrentHashMap 中的key和value都不能为空，会抛出NullPointerException

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
/** Implementation for put and putIfAbsent */
   final V putVal(K key, V value, boolean onlyIfAbsent) {
       if (key == null || value == null) throw new NullPointerException();
       int hash = spread(key.hashCode());
       int binCount = 0;
       for (Node<K,V>[] tab = table;;) {
           Node<K,V> f; int n, i, fh;
           if (tab == null || (n = tab.length) == 0)
           //初始化要考虑多线程
               tab = initTable();
               //tabAt,getObjectVolatile native线程安全
           else if ((f = tabAt(tab, i = (n - 1) & hash)) == null) {
               //使用cas，为空的时候U.compareAndSwapObject,比较交换，使用native 线程安全
               if (casTabAt(tab, i, null,
                            new Node<K,V>(hash, key, value, null)))
                   break;                   // no lock when adding to empty bin
           }
           else if ((fh = f.hash) == MOVED)
               tab = helpTransfer(tab, f);
           else {
               V oldVal = null;
               //直接锁数组上的Node
               //put数据的时候，在同一个数组下标下，锁，保证安全
               synchronized (f) {
                   if (tabAt(tab, i) == f) {
                      ...
                      ...
                   }
               }
               if (binCount != 0) {
                   if (binCount >= TREEIFY_THRESHOLD)
                       treeifyBin(tab, i);
                   if (oldVal != null)
                       return oldVal;
                   break;
               }
           }
       }
       //扩容要考虑其它线程
       addCount(1L, binCount);
       return null;

对比HashTable直接在方法上进行加锁，锁this

Hashtable上的put

1
2
3
4
5
6
7
8
9
10
public synchronized V put(K key, V value) {
    // Make sure the value is not null 不能为空
    if (value == null) {
        throw new NullPointerException();
    }
        // Makes sure the key is not already in the hashtable.
    Entry<?,?> tab[] = table;
    int hash = key.hashCode();//不能为空
    int index = (hash & 0x7FFFFFFF) % tab.length;
    @SuppressWarnings("unchecked")

Hashtable中的value不能为空，抛出异常NullPointerException；key不能为空，因为null不能获取hashcode();

这种特性，在分析源码的时候就直接出来的，不用背的，不用背的。

本文标题:Thinking in java 基础之HashMap

文章作者:Mu Xiaobai(慕小白)

发布时间:2019-10-17, 21:13:12

最后更新:2024-03-08, 10:29:56

原始链接:https://muxiaobai.github.io/2019/10/17/Thinking-in-java-基础之HashMap/

许可协议: "署名-非商用-相同方式共享 4.0" 转载请保留原文链接及作者。

文章目录
  1. 1. 一个简单的put引发的惨案
    1. 1.0.1. 怎么进行hash
    2. 1.0.2. 没有数组
    3. 1.0.3. 数组对应的位置为空
    4. 1.0.4. 数组对应的位置不为空，跟链表，超过TREEIFY_THRESHOLD时，当前hash对应的位置转为treeifyBin
    5. 1.0.5. 数组对应的位置不为空，跟红黑树
    6. 1.0.6. 超过最大值，要进行扩容
    7. 1.0.7. 扩容怎么判定原来的位置，怎么移动
    8. 1.0.8. ConCurrentHashMap是怎么加锁的
    9. 1.0.9. 对比HashTable直接在方法上进行加锁，锁this
,