HashMap原始碼解析——基於jdk11
阿新 • • 發佈:2018-11-07
HashMap原始碼解析——基於jdk11
插入流程
首先我們從put方法說起:
public V put(K key, V value) {
return putVal(hash(key), key, value, false, true);
}
繼續往下看putVal方法:
final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
boolean evict) {
//如果table為空就初始化一個table n為table的長度
Node<K,V>[] tab; Node<K,V> p; int n, i;
if ((tab = table) == null || (n = tab.length) == 0)
n = (tab = resize()).length;
//如果資料將要存放在陣列的位置為空 那麼就把該值的放在陣列上
if ((p = tab[ i = (n - 1) & hash]) == null)
tab[i] = newNode(hash, key, value, null);
else {
//如果key已經存在, e = 存在的值
Node<K,V> e; K k;
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
e = p;
else if (p instanceof TreeNode)
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
else {
for (int binCount = 0; ; ++binCount) {
//正常情況下,把值插入連結串列的最後面
if ((e = p.next) == null) {
p.next = newNode(hash, key, value, null);
//如果一個節點的元素太多 這個節點使用紅黑樹
if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
treeifyBin(tab, hash);
break;
}
//如果key已經存在, e = 存在的值
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
break;
p = e;
}
}
//e != null,說明存在相同的key
if (e != null) { // existing mapping for key
V oldValue = e.value;
//在一定條件下新的值可以覆蓋舊的值
if (!onlyIfAbsent || oldValue == null)
e.value = value;
afterNodeAccess(e);
return oldValue;
}
}
++modCount;
//資料超過一定的數量 擴容
if (++size > threshold)
resize();
afterNodeInsertion(evict);
return null;
}
獲取流程
首先看看
public V get(Object key) {
Node<K,V> e;
return (e = getNode(hash(key), key)) == null ? null : e.value;
}
然後我們再看
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 &&
//找到連結串列的頭
(first = tab[(n - 1) & hash]) != null) {
//如果連結串列頭就是要找的key 返回結果
if (first.hash == hash && // always check first node
((k = first.key) == key || (key != null && key.equals(k))))
return first;
//如果不是 就遍歷查詢連結串列
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;
}