java源码_HashMap(二)

增删改查方法

1.put,增,改

public V put(K key, V value) {
    return putVal(hash(key), key, value, false, true);
}

public V putIfAbsent(K key, V value) {
    return putVal(hash(key), key, value, true, true);
}

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)//空table则建table
        n = (tab = resize()).length;//建好table,n为table长度
    if ((p = tab[i = (n - 1) & hash]) == null)//hash之后的索引位置为空则直接放进去,p为索引为i的链表头
        tab[i] = newNode(hash, key, value, null);
    else {//不为空则插入

        Node<K,V> e; K k;
        if (p.hash == hash &&
            ((k = p.key) == key || (key != null && key.equals(k))))//如果链表头和要插入的节点相同
            e = p;
        else if (p instanceof TreeNode)//如果是红黑树则采用红黑树的put
            e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
        else {
            for (int binCount = 0; ; ++binCount) {//binCount为了统计是否到了变红黑树的阀值
                if ((e = p.next) == null) {//找到链表的尾节点,插入
                    p.next = newNode(hash, key, value, null);
                    if (binCount >= TREEIFY_THRESHOLD - 1) //如果超过变红黑树的阀值,变红黑树
                        treeifyBin(tab, hash);
                    break;
                }
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))//如果e和要插入的节点相同
                    break;
                p = e;
            }
        }
        if (e != null) { //如果e不为空,根据参数决定是否替换老的值
            V oldValue = e.value;
            if (!onlyIfAbsent || oldValue == null)
                e.value = value;
            afterNodeAccess(e);//对e进行操作后的函数,主要用于linkedHashMap实现LRU
            return oldValue;
        }
    }
    ++modCount;//结构性更改次数增加
    if (++size > threshold)//如果size超过阀值,resize
        resize();
    afterNodeInsertion(evict);
    return null;
}

2.remove,删

public boolean remove(Object key, Object value) {
    return removeNode(hash(key), key, value, true, true) != null;
}

public V remove(Object key) {
    Node<K,V> e;
    return (e = removeNode(hash(key), key, null, false, true)) == null ?
        null : e.value;
}

final Node<K,V> removeNode(int hash, Object key, Object value,
                           boolean matchValue, boolean movable) {
    Node<K,V>[] tab; Node<K,V> p; int n, index;
    if ((tab = table) != null && (n = tab.length) > 0 &&
        (p = tab[index = (n - 1) & hash]) != null) {//table不为空且table槽
        Node<K,V> node = null, e; K k; V v;
        if (p.hash == hash &&
            ((k = p.key) == key || (key != null && key.equals(k))))
            node = p;
        else if ((e = p.next) != null) {
            if (p instanceof TreeNode)
                node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
            else {
                do {
                    if (e.hash == hash &&
                        ((k = e.key) == key ||
                         (key != null && key.equals(k)))) {
                        node = e;
                        break;
                    }
                    p = e;
                } while ((e = e.next) != null);
            }
        }
        if (node != null && (!matchValue || (v = node.value) == value ||
                             (value != null && value.equals(v)))) {
            if (node instanceof TreeNode)
                ((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);
            else if (node == p)
                tab[index] = node.next;
            else
                p.next = node.next;
            ++modCount;
            --size;
            afterNodeRemoval(node);
            return node;
        }
    }
    return null;
}

3.get

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

final Node<K,V> getNode(int hash, Object key) {
    Node<K,V>[] tab; Node<K,V> first, e; int n; K k;
    if ((tab = table) != null && (n = tab.length) > 0 &&
        (first = tab[(n - 1) & hash]) != null) {//查询槽
        if (first.hash == hash && // always check first node
            ((k = first.key) == key || (key != null && key.equals(k))))//对比首节点
            return first;
        if ((e = first.next) != null) {
            if (first instanceof TreeNode)
                return ((TreeNode<K,V>)first).getTreeNode(hash, key);//如果是树节点,查询树节点
            do {//查询链表
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))
                    return e;
            } while ((e = e.next) != null);
        }
    }
    return null;
}