它是JDK11中LinkedHashMap中的死代码吗？

Question

我正在阅读 JDK 11 中的 LinkedHashMap 源代码，我发现了一段死代码（我不确定）

众所周知，LinkedHashMap 使用双向链表来保存所有元素的顺序。它有一个成员叫做 accessOrder

final boolean accessOrder;

默认情况下它是 false，但如果它被设置为 true，每次运行时get，它都会将它到达的元素移动到链表的末尾。这就是函数afterNodeAccess所做的。

//if accessOrder were set as true, after you visit node e, if e is not the end node of the linked list,
//it will move the node to the end of the linkedlist. 
    void afterNodeAccess(Node<K, V> e) {
        LinkedHashMap.Entry<K, V> last;

        if(accessOrder && (last = tail) != e) {

            //if enter `if` ?it indicates that e is not the end of the linked list, because (last=tail!=e)
            //then `a` as the after node of p(p is e after casting to LinkedHashMap.Entry) is never gonna be null. Only if p is last node of the linked list then a will be null.
            LinkedHashMap.Entry<K, V> p = (LinkedHashMap.Entry<K, V>) e, b = p.before, a = p.after;

            p.after = null;

            if(b == null) {
                head = a;
            } else {
                b.after = a;
            }

            // Is the if else clasue redundant? `a` must not be null.. the else clase will never be excuted.
            if(a != null) {
                a.before = b;
            } else {
                last = b;
            }

            if(last == null) {
                head = p;
            } else {
                p.before = last;
                last.after = p;
            }

            tail = p;

            ++modCount;
        }
    }

所以我的问题来了：

(accessOrder && (last = tail) != e意味着 e 不是链表的最后一个节点。如果 e 已经是最后一个节点，我们就不用做任何事情了吧？

那么a作为p的后节点，（p转换为LinkedHashMap.Entry后就是e），不能为空。只有当p是最后一个节点时才a可以为空。

那么下面这段代码有什么意义呢？

 // Is the if else clasue redundant? `a` must not be null.. the else clase will never be excuted.
            if(a != null) {
                a.before = b;
            } else {
                last = b;
            }

aalways != null，else 子句last = b永远不会被执行......那么它是死代码吗？

我也用accessorderset as做了一个实验true，然后我get是调试模式下的最后一个节点，似乎我永远无法进入上面的 else caluselast = b

有什么建议？

Answer 1

OP中提供的代码是单链表的节点移除算法，它将移除的节点设置为链表的尾部（重新定位到尾部）：

        LinkedHashMap.Entry<K, V> current = (LinkedHashMap.Entry<K, V>) e
        LinkedHashMap.Entry<K, V> pred = current.before, succ = current.after;

        current.after = null;

        // position the successor of the removed node correctly 
        // (either as the head of the list or as the successor of the node BEFORE the removed node)
        if(pred == null) {
            head = succ;
        } else {
            pred.after = succ ;
        }

        // position the predecessor of the removed node correctly
        // (either as the tail of the list or as the predecessor of the node AFTER the removed node)
        if(succ != null) {
            succ.before = pred;
        } else { // unreachable for non tail nodes
            last = pred;
        }

        // final step - if the predecessor of the removed node was null then the head
        // of the list is the removed node (the list contains a single node).
        // otherwise update the removed node as the tail of the list -
        // its predecessor will be the previous tail of the list
        if(last == null) { // unreachable for non tail nodes
            head = current;
        } else { 
            current.before = last;
            last.after = current;
        }

        tail = current;

该算法处理所有可能的情况，给定一个应该重新定位为链接情况尾部的节点。

在该方法的上下文中afterNodeAccess，一般情况算法中将存在一些冗余，因为由于 ，重新定位的节点永远不会位于列表的尾部(last = tail) != e。因此，更有效的算法是：

        current.after = null;
        // position the successor of the removed node correctly 
        // (either as the head of the list or as the successor of the node BEFORE the removed node)
        if(pred == null) {
            head = succ;
        } else {
            pred.after = succ ;
        }

        // position the predecessor of the removed node correctly
        // (as the predecessor of the node AFTER the removed node)
        // update the removed node as the tail of the list -
        // its predecessor will be the previous tail of the list
        succ.before = pred;
        current.before = last;
        last.after = current;
        tail = current;

正如holger在评论中提到的 - 这是一个经典的“复制粘贴”解决方案，恕我直言，它表明在某些情况下重用代码似乎效率低下且不清楚。

根据Johannes Kuhn的建议，您可以考虑向 OpenJDK 社区提交针对无法访问的代码的修复程序。请参阅有关如何完成此操作的参考资料。

参考：

• LinkedHashMap 的 AdoptOpen JDK 11 源代码
• 如何为 OpenJDK 提供修复