HashMap
JDK 1.8
构造方法
//最常用的构造方法,默认初始容量是16,懒加载
public HashMap() {
this.loadFactor = DEFAULT_LOAD_FACTOR; // 0.75
}
put方法
这个是关键!
public V put(K key, V value) {
return putVal(hash(key), key, value, false, true);
}
//先看下 hash(key) 里做了什么 👇
static final int hash(Object key) {
int h;
//这个hash值后续会参与对数组下标位置散列,即(n-1)%hash
//而数组容量通常不会很大,这意味着高位参与运算的概率小,为了散列分散更均匀
//我们将hash右移16位,与原hash相与,参与运算
return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}
//接下来看最重要的putVal
final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
boolean evict) {
Node<K,V>[] tab; Node<K,V> p; int n, i;
//如果第一次put,则创建table数组,懒加载,看① resize() 👇
if ((tab = table) == null || (n = tab.length) == 0)
n = (tab = resize()).length;
//散列求得位置,且该位置为空
if ((p = tab[i = (n - 1) & hash]) == null)
tab[i] = newNode(hash, key, value, null);
//这位置不为空
else {
Node<K,V> e; K k;
//这是key相同的情况,即覆盖情况
//看看如何判断key相等,先看hash相同吗?不同直接排除,再看地址或equals
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
e = p;
//红黑树的情况,不谈
else if (p instanceof TreeNode)
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
//该链表的头节点和key不相同,往下找
else {
for (int binCount = 0; ; ++binCount) {
//找到尾了还没找到,是在链表里新增一个值
//这里可以看出jdk1.8使用的是尾插法,因为在jdk1.7中使用头插法时,在多线程环境下会引发死链问题
if ((e = p.next) == null) {
p.next = newNode(hash, key, value, null);
//这里不一定会树化!看②处代码 👇
//所以链表树化的条件是:链表长度达到8且数组长度大于64(默认)
if (binCount >= TREEIFY_THRESHOLD - 1)
treeifyBin(tab, hash);
break;
}
//找到相同的key,是覆盖的情况
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
break;
p = e;
}
}
//给e赋值,这个e可能是新结点也可能是已存在覆盖的情况
if (e != null) { // existing mapping for key
V oldValue = e.value;
//onlyIfAbsent:key不存在才插入,相当于不能覆盖,只能新增
if (!onlyIfAbsent || oldValue == null)
e.value = value;
afterNodeAccess(e);
return oldValue;
}
}
//添加修改次数
++modCount;
//如果最后的容量超过阈值了,扩容
if (++size > threshold)
resize();
afterNodeInsertion(evict);
return null;
}
//① resize() 扩容
final Node<K,V>[] resize() {
Node<K,V>[] oldTab = table;
int oldCap = (oldTab == null) ? 0 : oldTab.length;
int oldThr = threshold;
int newCap, newThr = 0;
if (oldCap > 0) {
//已经最大了
if (oldCap >= MAXIMUM_CAPACITY) {
threshold = Integer.MAX_VALUE;
return oldTab;
}
//扩大成两倍
else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
oldCap >= DEFAULT_INITIAL_CAPACITY)
newThr = oldThr << 1; // double threshold
}
//加了初始值的构造函数懒加载
else if (oldThr > 0) // initial capacity was placed in threshold
newCap = oldThr;
//默认构造方法懒加载会到这个分支
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 < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?
(int)ft : Integer.MAX_VALUE);
}
threshold = newThr;
@SuppressWarnings({"rawtypes","unchecked"})
Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
table = newTab;
//rehash过程
if (oldTab != null) {
for (int j = 0; j < oldCap; ++j) {
Node<K,V> e;
//如果该桶不为空
if ((e = oldTab[j]) != null) {
oldTab[j] = null;
if (e.next == null)
newTab[e.hash & (newCap - 1)] = e;
else if (e instanceof TreeNode)
((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
else { // preserve order
链表hash...
}
}
}
}
return newTab;
//回去看putVal 👆
}
//② 树化
final void treeifyBin(Node<K,V>[] tab, int hash) {
int n, index; Node<K,V> e;
//如果数组为空或者数组长度大于64,就扩容
if (tab == null || (n = tab.length) < MIN_TREEIFY_CAPACITY)
resize();
else if ((e = tab[index = (n - 1) & hash]) != null) {
树化...
}
}
//回去看putVal 👆
get() 方法
get方法会简单些
public V get(Object key) {
Node<K,V> e;
return (e = getNode(hash(key), key)) == null ? null : e.value;
}
// 直接看getNode 👇
final Node<K,V> getNode(int hash, Object key) {
Node<K,V>[] tab; Node<K,V> first, e; int n; K k;
//数组不为空、长度大于0、hash求的位置上不为空,否则返回null
if ((tab = table) != null && (n = tab.length) > 0 &&
(first = tab[(n - 1) & hash]) != null) {
//和put一样了,先看hash,再看地址或equals
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);
do {
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
return e;
} while ((e = e.next) != null);
}
}
return null;
}
get还是比较简单的
remove() 方法
public V remove(Object key) {
Node<K,V> e;
return (e = removeNode(hash(key), key, null, false, true)) == null ?
null : e.value;
}
//直接看removeNode
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;
//一摸一样的if判断了,试试看能不能看懂
if ((tab = table) != null && (n = tab.length) > 0 &&
(p = tab[index = (n - 1) & hash]) != null) {
//这里的node就是要删除的结点
Node<K,V> node = null, e; K k; V v;
//先hash,再地址或equals 🤯
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
node = p;
//头结点不是,往下查
else if ((e = p.next) != null) {
if (p instanceof TreeNode)
node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
else {
//顺着链表查找了
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);
//是头节点
else if (node == p)
tab[index] = node.next;
else
p.next = node.next;
++modCount;
--size;
afterNodeRemoval(node);
return node;
}
//没找到要删除的
}
return null;
}
Q.E.D.