标签:nod java otto tin dentry initial hit previous access
之前hashmap处理冲突单纯使用链表法进行链接,1.8中,当一散列值中对应的链表长度超过8个后,会将链表转化为红黑树进行存储。
1.8中源码
public V put(K key, V value) { return putVal(hash(key), key, value, false, true); }
/** * Implements Map.put and related methods * * @param hash hash for key * @param key the key * @param value the value to put * @param onlyIfAbsent if true, don‘t change existing value * @param evict if false, the table is in creation mode. * @return previous value, or null if none */
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tab为空或长度为0,则分配内存resize() if ((tab = table) == null || (n = tab.length) == 0) n = (tab = resize()).length; //(n - 1) & hash找到put位置,如果为空,则直接put if ((p = tab[i = (n - 1) & hash]) == null) tab[i] = newNode(hash, key, value, null); else { Node<k,v> e; K k; //第一节节点hash值同,且key值与插入key相同 if (p.hash == hash &&((k = p.key) == key || (key != null && key.equals(k)))) e = p; else if (p instanceof TreeNode)//属于红黑树处理冲突 e = ((TreeNode<k,v>)p).putTreeVal(this, tab, hash, key, value); else { //链表处理冲突 for (int binCount = 0; ; ++binCount) { //p第一次指向表头,以后依次后移 if ((e = p.next) == null) { //e为空,表示已到表尾也没有找到key值相同节点,则新建节点 p.next = newNode(hash, key, value, null); //新增节点后如果节点个数到达阈值,则将链表转换为红黑树 if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st treeifyBin(tab, hash); break; } //容许null==null if (e.hash == hash &&((k = e.key) == key || (key != null && key.equals(k)))) break; p = e;//更新p指向下一个节点 } } //更新hash值和key值均相同的节点Value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; }
1.7源码
/** * Associates the specified value with the specified key in this map. * If the map previously contained a mapping for the key, the old * value is replaced. */ public V put(K key, V value) { if (table == EMPTY_TABLE) { inflateTable(threshold); } if (key == null) return putForNullKey(value); int hash = hash(key); int i = indexFor(hash, table.length); //这里的循环是关键 //当新增的key所对应的索引i,对应table[i]中已经有值时,进入循环体 for (Entry<K,V> e = table[i]; e != null; e = e.next) { Object k; //判断是否存在本次插入的key,如果存在用本次的value替换之前oldValue,相当于update操作 //并返回之前的oldValue if (e.hash == hash && ((k = e.key) == key || key.equals(k))) { V oldValue = e.value; e.value = value; e.recordAccess(this); return oldValue; } } //如果本次新增key之前不存在于HashMap中,modCount加1,说明结构改变了 modCount++; addEntry(hash, key, value, i); return null; }
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tab为空或长度为0,则分配内存resize()
if
((tab = table) ==
null
|| (n = tab.length) ==
0
)
n = (tab = resize()).length;
//(n - 1) & hash找到put位置,如果为空,则直接put
if
((p = tab[i = (n -
1
) & hash]) ==
null
)
tab[i] = newNode(hash, key, value,
null
);
else
{
Node<k,v> e; K k;
//第一节节点hash值同,且key值与插入key相同
if
(p.hash == hash &&((k = p.key) == key || (key !=
null
&& key.equals(k))))
e = p;
else
if
(p
instanceof
TreeNode)
//属于红黑树处理冲突
e = ((TreeNode<k,v>)p).putTreeVal(
this
, tab, hash, key, value);
else
{
//链表处理冲突
for
(
int
binCount =
0
; ; ++binCount) {
//p第一次指向表头,以后依次后移
if
((e = p.next) ==
null
) {
//e为空,表示已到表尾也没有找到key值相同节点,则新建节点
p.next = newNode(hash, key, value,
null
);
//新增节点后如果节点个数到达阈值,则将链表转换为红黑树
if
(binCount >= TREEIFY_THRESHOLD -
1
)
// -1 for 1st
treeifyBin(tab, hash);
break
;
}
//容许null==null
if
(e.hash == hash &&((k = e.key) == key || (key !=
null
&& key.equals(k))))
break
;
p = e;
//更新p指向下一个节点
}
}
//更新hash值和key值均相同的节点Value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
;
}
标签:nod java otto tin dentry initial hit previous access
原文地址:http://www.cnblogs.com/aksdenjoy/p/6548783.html