HashMap 源码分析

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

/**
 * Basic hash bin node, used for most entries.  (See below for
 * TreeNode subclass, and in LinkedHashMap for its Entry subclass.)
 */
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;
    }
}

final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
               boolean evict) {
    Node<K,V>[] tab; Node<K,V> p; int n, i;
    //将table赋值给tab，如果table为空或者长度为0，则调用resize()初始化，再把长度赋值给n
    if ((tab = table) == null || (n = tab.length) == 0)
        n = (tab = resize()).length;
    //这里(n - 1) & hash相当于hash%(n-1)，判断tab[i]这个区域是否已经有值，如果没有，则新建一个节点，并且赋值给p，把节点放到这个区域
    if ((p = tab[i = (n - 1) & hash]) == null)
        tab[i] = newNode(hash, key, value, null);
    //如果tab[i]里面已经存在值了
    else {
        Node<K,V> e; K k;
        //如果p这个节点的传进来的是一样的，则把p赋值给e
        if (p.hash == hash &&
            ((k = p.key) == key || (key != null && key.equals(k))))
            e = p;
        //如果p为树类型节点，则调用树的putTreeVal方法，此方法就是一个红黑树添加
        else if (p instanceof TreeNode)
            e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
        //遍历区域中所有节点
        else {
            for (int binCount = 0; ; ++binCount) {
                //将p的子节点赋值给e，如果区域中没有相同key的节点则直接插入到区域中（可能是链表可能是红黑树，具体查看treeifyBin方法），退出循环
                if ((e = p.next) == null) {
                    p.next = newNode(hash, key, value, null);
                    if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                        treeifyBin(tab, hash);
                    break;
                }
                //如果区域中有相同key的节点直接退出循环，
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))
                    break;
                //以上都没有，则e赋值给p继续循环
                p = e;
            }
        }
        //e不为空，说明之前区域中存在key与现在新的key相同，这时候将新值覆盖原来的旧值，并且返回旧值
        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;
}

public V get(Object key) {
    Node<K,V> e;
    return (e = getNode(hash(key), key)) == null ? null : e.value;
}