Java Jdk1.8 HashMap源代码阅读笔记二
三、源代码阅读
3、元素包括containsKey(Object key)
/**
* Returns <tt>true</tt> if this map contains a mapping for the
* specified key.
*
* @param key The key whose presence in this map is to be tested
* @return <tt>true</tt> if this map contains a mapping for the specified
* key.
*/
public boolean containsKey(Object key) {
return getNode(hash(key), key) != null;
}
/**
* Implements Map.get and related methods
*
* @param hash hash for key
* @param key the key
* @return the node, or null if none
*/
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 &&
//推断tab[hash]位置是否有值
(first = tab[(n - 1) & hash]) != null) {
if (first.hash == hash && // always check first node
((k = first.key) == key || (key != null && key.equals(k))))
return first;
if ((e = first.next) != null) {
if (first instanceof TreeNode)
return ((TreeNode<K,V>)first).getTreeNode(hash, key);
//遍历寻找
do {
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
return e;
} while ((e = e.next) != null);
}
}
return null;
}
/**
* Calls find for root node.
*/
final TreeNode<K,V> getTreeNode(int h, Object k) {
return ((parent != null) ? root() : this).find(h, k, null);
}
/**
* Returns root of tree containing this node.
* 获取红黑树的根
*/
final TreeNode<K,V> root() {
for (TreeNode<K,V> r = this, p;;) {
if ((p = r.parent) == null)
return r;
r = p;
}
}
/**
* Finds the node starting at root p with the given hash and key.
* The kc argument caches comparableClassFor(key) upon first use
* comparing keys.
*/
final TreeNode<K,V> find(int h, Object k, Class<?> kc) {// k即key，kc为null
TreeNode<K,V> p = this;
do {
int ph, dir; K pk;
TreeNode<K,V> pl = p.left, pr = p.right, q;
if ((ph = p.hash) > h)// ph存当前节点hash
p = pl;
else if (ph < h) // 所查hash比当前节点hash大
p = pr;// 查右子树
else if ((pk = p.key) == k || (k != null && k.equals(pk)))
return p;// hash、key均同样。【找到了！】返回当前节点
else if (pl == null)// hash等，key不等，且当前节点的左节点null
p = pr;//查右子树
else if (pr == null)
p = pl;
//get->getTreeNode传递的kc为null。||逻辑或,短路运算,有真就可以
// false || (false && ？？)
else if ((kc != null ||
(kc = comparableClassFor(k)) != null) &&
(dir = compareComparables(kc, k, pk)) != 0)
p = (dir < 0) ? pl : pr;
else if ((q = pr.find(h, k, kc)) != null)
return q;
else
p = pl;
} while (p != null);
return null;
}
4、get(Object key)
/**
* 返回value或null
*/
public V get(Object key) {
Node<K,V> e;
return (e = getNode(hash(key), key)) == null ? null : e.value;
}
函数getNode在阅读笔记一中已经记录。
5、移除 remove(Object key)
/**
* Removes the mapping for the specified key from this map if present.
*
* @param key key whose mapping is to be removed from the map
* @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 remove(Object key) {
Node<K,V> e;
return (e = removeNode(hash(key), key, null, false, true)) == null ?
null : e.value;
}
/**
* Implements Map.remove and related methods
*
* @param hash hash for key
* @param key the key
* @param value the value to match if matchValue, else ignored
* @param matchValue if true only remove if value is equal
* @param movable if false do not move other nodes while removing
* @return the node, or null if none
*/
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) {
Node<K,V> node = null, e; K k; V v;
if (p.hash == hash &&
//先比較内存地址。假设地址不一致。再调用equals进行比較
((k = p.key) == key || (key != null && key.equals(k))))
node = p;
else if ((e = p.next) != null) {
//假设是以红黑树处理冲突。则通过getTreeNode查找
if (p instanceof TreeNode)
node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
else {
//假设是以链式的方式处理冲突。则通过遍历链表来寻找节点
do {
if (e.hash == hash &&
((k = e.key) == key ||
(key != null && key.equals(k)))) {
node = e;
break;
}
p = e;
} while ((e = e.next) != null);
}
}
//比对找到的key的value跟要删除的是否匹配
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;
//已从结构上改动 此列表的次数
++modCount;
--size;
//回调
afterNodeRemoval(node);
return node;
}
}
return null;
}
/**
* Removes the given node, that must be present before this call.
* This is messier than typical red-black deletion code because we
* cannot swap the contents of an interior node with a leaf
* successor that is pinned by "next" pointers that are accessible
* independently during traversal. So instead we swap the tree
* linkages. If the current tree appears to have too few nodes,
* the bin is converted back to a plain bin. (The test triggers
* somewhere between 2 and 6 nodes, depending on tree structure).
*/
final void removeTreeNode(HashMap<K,V> map, Node<K,V>[] tab,
boolean movable) {
int n;
if (tab == null || (n = tab.length) == 0)
return;
int index = (n - 1) & hash;
TreeNode<K,V> first = (TreeNode<K,V>)tab[index], root = first, rl;
TreeNode<K,V> succ = (TreeNode<K,V>)next, pred = prev;
if (pred == null)
tab[index] = first = succ;
else
pred.next = succ;
if (succ != null)
succ.prev = pred;
if (first == null)
return;
if (root.parent != null)
root = root.root();
if (root == null || root.right == null ||
(rl = root.left) == null || rl.left == null) {
tab[index] = first.untreeify(map); // too small
return;
}
TreeNode<K,V> p = this, pl = left, pr = right, replacement;
if (pl != null && pr != null) {
TreeNode<K,V> s = pr, sl;
while ((sl = s.left) != null) // find successor
s = sl;
boolean c = s.red; s.red = p.red; p.red = c; // swap colors
TreeNode<K,V> sr = s.right;
TreeNode<K,V> pp = p.parent;
if (s == pr) { // p was s's direct parent
p.parent = s;
s.right = p;
}
else {
TreeNode<K,V> sp = s.parent;
if ((p.parent = sp) != null) {
if (s == sp.left)
sp.left = p;
else
sp.right = p;
}
if ((s.right = pr) != null)
pr.parent = s;
}
p.left = null;
if ((p.right = sr) != null)
sr.parent = p;
if ((s.left = pl) != null)
pl.parent = s;
if ((s.parent = pp) == null)
root = s;
else if (p == pp.left)
pp.left = s;
else
pp.right = s;
if (sr != null)
replacement = sr;
else
replacement = p;
}
else if (pl != null)
replacement = pl;
else if (pr != null)
replacement = pr;
else
replacement = p;
if (replacement != p) {
TreeNode<K,V> pp = replacement.parent = p.parent;
if (pp == null)
root = replacement;
else if (p == pp.left)
pp.left = replacement;
else
pp.right = replacement;
p.left = p.right = p.parent = null;
}
TreeNode<K,V> r = p.red ? root : balanceDeletion(root, replacement);
if (replacement == p) { // detach
TreeNode<K,V> pp = p.parent;
p.parent = null;
if (pp != null) {
if (p == pp.left)
pp.left = null;
else if (p == pp.right)
pp.right = null;
}
}
if (movable)
moveRootToFront(tab, r);
}
方法 final void removeTreeNode临时没有吃透，待兴许补充。
四、小结
HashMap高性能须要下面几点：
1、高效的hash算法
2、保证hash值到内存地址（数组索引）的映射速度
3、依据内存地址（数组索引）能够直接得到对应的值
參考文章：https://yq.aliyun.com/articles/36812?
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作者：jiankunking 出处：http://blog.csdn.net/jiankunking
