【发布时间】:2019-10-10 23:49:03
【问题描述】:
我只是想向 C 中的排序双向链表插入一个值。 当我使用以下打印出来时,它永远不会显示新创建的节点。我创建了一个新节点,然后设置值,然后更新新节点和它前面的节点的 prev 和 next 指针。我确定这与通过引用传递有关,并想了解原因?
struct NodeType {
int data;
struct NodeType * prev;
struct NodeType * next;
}*head, *last;
void insert_double(int key);
void displayList();
void find_node(int key);
int main()
{
head = NULL;
last = NULL;
/* Create a list with one as value and set as head */
head = (struct NodeType *)malloc(sizeof(struct NodeType));
head->data = 3;
head->prev = NULL;
head->next = NULL;
last = head;
int value=1;
insert_double(value);
printf("0\n");
displayList();
printf("1\n");
value=2;
printf("2\n");
find_node(value);
printf("3\n");
displayList();
return 0;
}
void displayList()
{
struct NodeType * temp;
int n = 1;
if(head == NULL)
{
printf("List is empty.\n");
}
else
{
temp = head;
printf("DATA IN THE LIST:\n");
while(temp != NULL)
{
printf("DATA of %d node = %d\n", n, temp->data);
n++;
/* Move the current pointer to next node */
temp = temp->next;
}
}
}
void find_node(int key)
{
struct NodeType * temp;
struct NodeType * newnode;
newnode->data=key;
if(head == NULL){
printf("No nodes");
}
else{
temp=head;
while(temp!=NULL)
{
if((temp->data)< key){
newnode->prev=temp->prev;
newnode->next=temp;
temp->prev=newnode;
break;
}
else{
temp=temp->next;
}
}
}
}
void insert_double(int key)
{
struct NodeType * newnode;
if(head == NULL)
{
printf("Empty!\n");
}
else
{
newnode = (struct NodeType *)malloc(sizeof(struct NodeType));
newnode->data = key;
newnode->next = head; // Point to next node which is currently head
newnode->prev = NULL; // Previous node of first node is NULL
/* Link previous address field of head with newnode */
head->prev = newnode;
/* Make the new node as head node */
head = newnode;
}
}
【问题讨论】:
-
在 find_node() 中 new_node 被使用(通过 new_node->data = key),然后 new_node 被赋值。
标签: c doubly-linked-list