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The code will crash spectacularly. If p->next is NULL, then p->next->prev will give you a null pointer exception. The code is meaningless. There must be typos.
Choose the correct alternative to replace the blank line. (A) q = NULL; p->next = head; head = p; (B) q->next = NULL; head = p; p->next = head; (C) head = p; p->next = q; q->next = NULL; (D) q->next = NULL; p->next = head; head = p; Answer (D) When the while loop ends, q contains address of second last node and p contains address of last node. So we need to do following things after while loop ...
p->next=NULL:p指针指向的对象的next属性为空r->next=p:r指针指向的对象的下一个为p指向的对象r=p：r指针指向p指针指向的对象。
文章浏览阅读1.6w次，点赞87次，收藏274次。本文详细解析了C语言中结构体与指针的概念，特别是p->next的操作含义，并通过具体实例展示了线性表插入操作中的指针交换过程。
either cause a null pointer dereference or append list m to the end of list n
文章浏览阅读1.6w次，点赞44次，收藏136次。本文通过两个具体的while循环示例，详细解析了链表的遍历过程，包括如何访问链表中的每个节点并打印其数据。
It means that the data stored in the address of p pointer must be smaller or equal to the data in the next node i.e p->next->data. Eg- 2->2-.3->4 which is in non decreasing order and also more valid than option A.
CASE 1: When lists are not empty, so according to the code given, join will append list m to end of the list n. Diagram CASE 2: When lists are empty but valid like this, node *head; head - > NULL; p = head p- > next! = NULL. So, it will create a NULL reference issue. So, answer will be either cause a null pointer dereference or append list m to ...
p!= NULL 还是 P->next!=NULL，代码先锋网，一个为软件开发程序员提供代码片段和技术文章聚合的网站。
At each iteration of the loop, p will point to each node in the list in turn: for (Node p = head; p != null; p = p.next ) { // Do something with each node, such as print out the item } OR use a while loop: Node p = head; while (p != null) { // Do something with each node, such as print out the item p = p.next; }