remove操作非常类似put，但要注意一点区别，中间那个for循环是做什么用的呢？从代码来看，就是将定位之后的所有entry克隆并拼回前面去，但有必要吗？每次删除一个元素就要将那之前的元素克隆一遍？这点其实是由entry的不变性来决定的，仔细观察entry定义，发现除了value，其他所有属性都是用final来修饰的，这意味着在第一次设置了next域之后便不能再改变它，取而代之的是将它之前的节点全都克隆一次。
public V remove(Object key) { return replaceNode(key, null, null); } /** * Implementation for the four public remove/replace methods: * Replaces node value with v, conditional upon match of cv if * non-null. If resulting value is null, delete. */ final V replaceNode(Object key, V value, Object cv) { int hash = spread(key.hashCode()); for (Node<K,V>[] tab = table;;) { Node<K,V> f; int n, i, fh; if (tab == null || (n = tab.length) == 0 || (f = tabAt(tab, i = (n - 1) & hash)) == null) break; else if ((fh = f.hash) == MOVED) tab = helpTransfer(tab, f); else { V oldVal = null; boolean validated = false; synchronized (f) { if (tabAt(tab, i) == f) { if (fh >= 0) { validated = true; for (Node<K,V> e = f, pred = null;;) { K ek; if (e.hash == hash && ((ek = e.key) == key || (ek != null && key.equals(ek)))) { V ev = e.val; if (cv == null || cv == ev || (ev != null && cv.equals(ev))) { oldVal = ev; if (value != null) e.val = value; else if (pred != null) pred.next = e.next; else setTabAt(tab, i, e.next); } break; } pred = e; if ((e = e.next) == null) break; } } else if (f instanceof TreeBin) { validated = true; TreeBin<K,V> t = (TreeBin<K,V>)f; TreeNode<K,V> r, p; if ((r = t.root) != null && (p = r.findTreeNode(hash, key, null)) != null) { V pv = p.val; if (cv == null || cv == pv || (pv != null && cv.equals(pv))) { oldVal = pv; if (value != null) p.val = value; else if (t.removeTreeNode(p)) setTabAt(tab, i, untreeify(t.first)); } } } } } if (validated) { if (oldVal != null) { if (value == null) addCount(-1L, -1); return oldVal; } break; } } } return null; }