hashMap的线程不安全

/**
     * Basic hash bin node, used for most entries.  (See below for
     * TreeNode subclass, and in LinkedHashMap for its Entry subclass.)
     */
    static class Node implements Map.Entry {
        final int hash;
        final K key;
        V value;
        Node next;

        Node(int hash, K key, V value, Node 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;
        }
    }

final int hash;
final K key;
V value;
Node next;

/**
     * 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[] tab; Node p; int n, i;
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;
        if ((p = tab[i = (n - 1) & hash]) == null)
            //1位置 此位置为空，则直接创建newNode
            tab[i] = newNode(hash, key, value, null);
        else {
        //如果此处有元素
            Node e; K k;
            //如果此处有元素，且key值相同，则覆盖元素
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;
                //如果当前位置的元素的结构是树节点，则将待添加的数据作为树节点，添加到红黑树结构中
            else if (p instanceof TreeNode)
                e = ((TreeNode)p).putTreeVal(this, tab, hash, key, value);
                //剩余情况，将新元素追加到当前链表元素的next节点
            else {
                for (int binCount = 0; ; ++binCount) {
                    if ((e = p.next) == null) {
                        p.next = newNode(hash, key, value, null);
                        //如果链表长度达到8，要将链表结构转化为红黑树
                        if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                            treeifyBin(tab, hash);
                        break;
                    }
                    //如果发现相同的key，同样覆盖
                    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[] resize() {
        //旧数据
        Node[] oldTab = table;
        //旧map长度
        int oldCap = (oldTab == null) ? 0 : oldTab.length;
        //要调整大小的下一个值
        int oldThr = threshold;
        int newCap, newThr = 0;
        if (oldCap > 0) {
            //如果当前数组长度大于等于最大值（1 <<30），就把threshold调到最大，无法再创建更大的数组了，直接返回原有数据
            if (oldCap >= MAXIMUM_CAPACITY) {
                threshold = Integer.MAX_VALUE;
                return oldTab;
            }
            //如果当前长度*2 小于最大值，并且当前长度大于等于默认长度16，阈值*2
            else if ((newCap = oldCap <<1) 
                     oldCap >= DEFAULT_INITIAL_CAPACITY)
                newThr = oldThr <<1; // double threshold
        }
        //如果阈值大于0，则将新数组的长度设置为阈值=16
        else if (oldThr > 0) // initial capacity was placed in threshold
            newCap = oldThr;
           // 零初始阈值表示使用默认值
        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 float)MAXIMUM_CAPACITY ?
                      (int)ft : Integer.MAX_VALUE);
        }
        threshold = newThr;
        @SuppressWarnings({"rawtypes","unchecked"})
        //创建新的数组
            Node[] newTab = (Node[])new Node[newCap];
        table = newTab;
        //将旧数据迁移到新数组
        if (oldTab != null) {
            for (int j = 0; j j) {
                Node e;
                if ((e = oldTab[j]) != null) {
                    oldTab[j] = null;
                    if (e.next == null)
                        //为每个不为空的key重新计算hash值 
                        newTab[e.hash & (newCap - 1)] = e;
                        //处理红黑树中数据
                    else if (e instanceof TreeNode)
                        ((TreeNode)e).split(this, newTab, j, oldCap);
                        //处理链表数据
                    else { // preserve order
                        Node loHead = null, loTail = null;
                        Node hiHead = null, hiTail = null;
                        Node 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;
    }