它是JDK11中LinkedHashMap中的死代码吗？

java java-8 java-11 dead-code linkedhashmap

2022-09-03 08:02:37

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使用双链表来保留所有元素的顺序。它有一个名为accessOrder

final boolean accessOrder;

默认情况下，它是 false，但如果设置为 true，则每次运行时，它都会将它所得到的元素移动到链表的末尾。这就是该函数的作用。getafterNodeAccess

//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表示 e 不是链表的最后一个节点。如果e已经是最后一个节点，我们不必做任何正确的事情？

然后作为 p 的后节点（p 在转换为 LinkedHashMap.Entry 后为 e），它不能为空。仅当是最后一个节点时，才能为空。apa

那么下面的代码段有什么意义呢？

 // 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;
            }

a总是，else 子句将永远不会被执行。...那么它是死代码吗？!= nulllast = b

另外，我用设置为做一个实验，然后我是调试模式下的最后一个节点，似乎我永远无法进入上述其他caluseaccessordertruegetlast = b

有什么建议吗？

答案 1

OP 中提供的代码是单个链表的节点删除算法，它将删除的节点设置为列表的尾部（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;

此算法处理给定一个节点的所有可能情况，该节点应重新定位为链接情况的尾部。

在该方法的上下文中，在一般情况下算法中会有一些冗余，因为由于 .因此，更有效的算法是：afterNodeAccess(last = tail) != e

        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，在某些情况下重用代码似乎是低效和不清楚的。

正如Johannes Kuhn所建议的那样，您可以考虑向OpenJDK社区提交无法访问的代码的修复程序。请参阅有关如何执行此操作的参考资料。

引用：

答案 2