問題描述

public V put(K key, V value) {if (key == null) return putForNullKey(value);int hash = hash(key);int i = indexFor(hash, table.length);for (Entry<K,V> e = table[i]; e != null; e = e.next) { Object k; if (e.hash == hash && ((k = e.key) == key || key.equals(k))) {V oldValue = e.value;e.value = value;e.recordAccess(this);return oldValue; }}modCount++;addEntry(hash, key, value, i);return null; }

void addEntry(int hash, K key, V value, int bucketIndex) { if ((size >= threshold) && (null != table[bucketIndex])) {resize(2 * table.length);hash = (null != key) ? hash(key) : 0;bucketIndex = indexFor(hash, table.length); } createEntry(hash, key, value, bucketIndex);}

void createEntry(int hash, K key, V value, int bucketIndex) {Entry<K,V> e = table[bucketIndex];table[bucketIndex] = new Entry<>(hash, key, value, e);size++; }

如果put 鍵索引相同key不同的元素，addEntry()怎么保存元素到鏈表尾部，createEntry()方法？

問題解答

看這段源碼

final V putVal(int hash, K key, V value, boolean onlyIfAbsent, boolean evict) {Node<K,V>[] tab; Node<K,V> p; int n, i;if ((tab = table) == null || (n = tab.length) == 0) n = (tab = resize()).length;if ((p = tab[i = (n - 1) & hash]) == null) tab[i] = newNode(hash, key, value, null);else { 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) { if ((e = p.next) == null) {p.next = newNode(hash, key, value, null);if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st treeifyBin(tab, hash);break; } if (e.hash == hash &&((k = e.key) == key || (key != null && key.equals(k))))break; p = e;} } if (e != null) { // existing mapping for keyV oldValue = e.value;if (!onlyIfAbsent || oldValue == null) e.value = value;afterNodeAccess(e);return oldValue; }}++modCount;if (++size > threshold) resize();afterNodeInsertion(evict);return null; }

如果相同key，不同value會進入這段：

if (e != null) { // existing mapping for keyV oldValue = e.value;if (!onlyIfAbsent || oldValue == null) e.value = value;afterNodeAccess(e);return oldValue; }