HashMap的实现
今天复习了HashMap，就和大家分享一下：
1.HashMap源码分析
- 初始化容量
/**
* The default initial capacity - MUST be a power of two.
*/
static final int DEFAULT_INITIAL_CAPACITY = 1 << 4; // aka 16初始化容量为1 << 4即16
- 最大容量
/**
* The maximum capacity, used if a higher value is implicitly specified
* by either of the constructors with arguments.
* MUST be a power of two <= 1<<30.
*/
static final int MAXIMUM_CAPACITY = 1 << 30;最大容量为2^30
- 装载因子
/**
* The load factor used when none specified in constructor.
*/
static final float DEFAULT_LOAD_FACTOR = 0.75f;默认装载因子是0.75，即在容量使用3/4是进行扩容
- 由链表转换成树的阈值TREEIFY_THRESHOLD
/**
* The bin count threshold for using a tree rather than list for a
* bin. Bins are converted to trees when adding an element to a
* bin with at least this many nodes. The value must be greater
* than 2 and should be at least 8 to mesh with assumptions in
* tree removal about conversion back to plain bins upon
* shrinkage.
*/
static final int TREEIFY_THRESHOLD = 8;当数组上的链表的长度大于8时，由链表转换成红黑树，这也是在jdk8中新增加的
- 由树转换成链表的阈值UNTREEIFY_THRESHOLD
/**
* The bin count threshold for untreeifying a (split) bin during a
* resize operation. Should be less than TREEIFY_THRESHOLD, and at
* most 6 to mesh with shrinkage detection under removal.
*/
static final int UNTREEIFY_THRESHOLD = 6;当数组上的链表的长度小于6时，由红黑树转换为链表
- MIN_TREEIFY_CAPACITY
/**
* The smallest table capacity for which bins may be treeified.
* (Otherwise the table is resized if too many nodes in a bin.)
* Should be at least 4 * TREEIFY_THRESHOLD to avoid conflicts
* between resizing and treeification thresholds.
*/
static final int MIN_TREEIFY_CAPACITY = 64;被树化时最小的hash表容量
- put方法
/**
* 计算key的hash值
*/
static final int hash(Object key) {
int h;
return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}
/**
* 插入key-value
*/
public V put(K key, V value) {
return putVal(hash(key), key, value, false, true);
}
/**
* 实现put操作
*
* @param key的hash值
* @param key值
* @param value值
* @param onlyIfAbsent if true, don't change existing value
* @param evict if false, the table is in creation mode.
* @return 之前的value
*/
final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
boolean evict) {
Node<K,V>[] tab; Node<K,V> p; int n, i;
//1. 如果当前table为空，新建默认大小的table
if ((tab = table) == null || (n = tab.length) == 0)
n = (tab = resize()).length;
//2. 获取当前key对应的节点
if ((p = tab[i = (n - 1) & hash]) == null)
//3. 如果不存在，新建节点
tab[i] = newNode(hash, key, value, null);
else {
//4. 存在节点
Node<K,V> e; K k;
//5. key的hash相同，key的引用相同或者key equals，则覆盖
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
e = p;
//6. 如果当前节点是一个红黑树树节点，则添加树节点
else if (p instanceof TreeNode)
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
//7. 不是红黑树节点，也不是相同节点，则表示为链表结构
else {
for (int binCount = 0; ; ++binCount) {
//8. 找到最后那个节点
if ((e = p.next) == null) {
p.next = newNode(hash, key, value, null);
//9. 如果链表长度超过8转成红黑树
if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
treeifyBin(tab, hash);
break;
}
//10.如果链表中有相同的节点，则覆盖
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;
//是否替换掉value值
if (!onlyIfAbsent || oldValue == null)
e.value = value;
afterNodeAccess(e);
return oldValue;
}
}
//记录修改次数
++modCount;
//是否超过容量，超过需要扩容
if (++size > threshold)
resize();
afterNodeInsertion(evict);
return null;
}