HashMap源码分析

源码分析

一.construct

    public HashMap(int initialCapacity, float loadFactor) {
        //valid loadFactor ...
        this.loadFactor = loadFactor;
        //计算容量
        this.threshold = tableSizeFor(initialCapacity);
    }

二.tableSizeFor

    /**
     * Returns a power of two size for the given target capacity.
     */
    static final int tableSizeFor(int cap) {
           //排查已经是2^n
        int n = cap - 1;
        n |= n >>> 1;
        n |= n >>> 2;
        n |= n >>> 4;
        n |= n >>> 8;
        n |= n >>> 16;
          //分析
          /**
         * 0000 001x xxxx xxxx  //假设任意二进制数,查找比它大,且是2^n的数
         * 0000 0100 0000 0000  //一定会得到这个值,以为每个位数都是2^n
         * n|n>>>1
         * n|n>>>2
         * n|n>>>4
         * n|n>>>8
         * n|n>>>16             //保证所有的数字都被覆盖
         * 最终结果
         * 0000 0111 1111 1111  //+1即可
         */

        return (n < 0) ? 1 : (n >= MAXIMUM_CAPACITY) ? MAXIMUM_CAPACITY : n + 1;
    }

三.put

    //1
    public V put(K key, V value) {
        return putVal(hash(key), key, value, false, true);
    }
    //2
    static final int hash(Object key) {
        int h;
         /**
         * 扰动函数
         * 包含高位地位的信息 -减少碰撞
         * 0abc defg hijk lmno
         * 0000 0000 0abc defg
         * ---或运算--
         */
        return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
    }
    //3
    final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
                   boolean evict) {
        Node<K,V>[] tab; Node<K,V> p; int n, i;
          // 如果桶容量为0,则初始化
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;
        //如果根据hash计算下标的位置没有元素则直接放置该位置
        //(n-1)&hash  定位
          //(2n-1)&hash 扩容重新定位(所以resize时,要么原位置,要么要位置2倍)
        //
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);
        else {
            Node<K,V> e; K k;
            //如果桶中第一个元素与插入元素key的hash相同,则将该点保存在e中
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;
            else if (p instanceof TreeNode)
              //如果是树,则putTreeVal
                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
            else {
                for (int binCount = 0; ; ++binCount) {
                    if ((e = p.next) == null) {
                       //如果遍历完毕,key不存在,则插入链表最后一个节点
                        p.next = newNode(hash, key, value, null);
                        if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                          //树化
                            treeifyBin(tab, hash);
                        break;
                    }
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        break;
                    p = e;
                }
            }
              //如果找到了key对应的元素
            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;
    }

先读这么多吧...