之前一直有这个疑问,今天有时间就把源码看了看终于知道了原理。分享给大家也做笔记自己可以随后查看。
linkedHashMap entry 集成了hashMap.Entry 然后定义了一个before与after用来存储当前key的上一个值引用和下一个值的引用。然后再addBefore的适合,把上一个值设置成当前对象引用。因为现在还不知道下一个是什么,所以默认也是当前对象引用。
*/ private static class Entry<K,V> extends HashMap.Entry<K,V> { // These fields comprise the doubly linked list used for iteration. Entry<K,V> before, after; Entry(int hash, K key, V value, HashMap.Entry<K,V> next) { super(hash, key, value, next); } /** * Inserts this entry before the specified existing entry in the list. */ private void addBefore(Entry<K,V> existingEntry) { after = existingEntry; before = existingEntry.before; before.after = this; after.before = this; }
然后linkedHashMap 定义了一个header变量用来存储第一个值的引用地址,这样迭代的时候就先迭代header,然后根据header 的next找到下一个迭代对象。
private transient Entry<K,V> header;
然后通过LinkedHashMapIterator的nextEntry方法就能看出nextEntry = e.after。 也就是说nextEntry是上一个值的下一个值。上一个值是当前对象,下一个值也就等于下一个对象,所以这就是它保持有序的原理。
private abstract class LinkedHashIterator<T> implements Iterator<T> { Entry<K,V> nextEntry = header.after; Entry<K,V> lastReturned = null; /** * The modCount value that the iterator believes that the backing * List should have. If this expectation is violated, the iterator * has detected concurrent modification. */ int expectedModCount = modCount; public boolean hasNext() { return nextEntry != header; } public void remove() { if (lastReturned == null) throw new IllegalStateException(); if (modCount != expectedModCount) throw new ConcurrentModificationException(); LinkedHashMap.this.remove(lastReturned.key); lastReturned = null; expectedModCount = modCount; } Entry<K,V> nextEntry() { if (modCount != expectedModCount) throw new ConcurrentModificationException(); if (nextEntry == header) throw new NoSuchElementException(); Entry<K,V> e = lastReturned = nextEntry; nextEntry = e.after; return e; } }
简单来说,就是第一个赋值是给header,第二次赋值是给header.after。
然后每次迭代就通过after知道了下一个值,也就保持了有序。
时间: 2024-10-05 05:07:36