链表中显示节点的代码不起作用

Question

我写了链表插入的代码。但是当我运行这段代码时，我的显示功能无法正常工作并打印其他内容。

**ERROR IN CODE**

在这里，我将head函数声明为全局变量，并将其设置为NULL。在那之后，当我 return 进入主要功能并通过它时。在一些代码将一个节点分配给结构 p 并将第一个节点声明为 head 之后（现在我们的 head 不是 NULL，它包含节点中的一些值），然后我们调用显示函数。现在当我们return到显示函数和运行它时，它显示我们的头是NULL而不是NULL。

下面是我的代码

#include <stdio.h>
#include <stdlib.h>

struct node
{
    int data;
    struct node* next;
};
struct node* head = NULL, * tail = NULL;

void printList()
{
    struct node* ptr = head;
    if (head == NULL)
    {
        printf("Empty list\n");
    }
    printf("Nodes of the list are: \n");
    while (ptr != NULL)
    {
        printf("%d \t", ptr->data);
        ptr = ptr->next;
    }
    printf("\n");
}

struct node* insertion(struct node** head, struct node** tail, int ele, int pos)
{
    int i;
    struct node* new_node, * temp;
    new_node = (struct node*)malloc(sizeof(struct node));
    if (new_node == NULL)
    {
        printf(" Dynamic memory allocation failed. Insertion can not be done");
        return NULL;
    }
    else
    {
        new_node->data = ele;
        if (pos < 1)
        {
            printf(" Inappropriate position");
            return NULL;
        }
        else if (*head == NULL)
        {
            new_node->next = NULL;
            *head = *tail = new_node;
            return (new_node);
        }
        else if (pos == 1)
        {
            new_node->next = *head;
            *head = new_node;
            return(new_node);
        }
        else
        {
            temp = *head;
            i = 1;
            while (i < pos - 1 && temp != NULL)
            {
                temp = temp->next;
                i++;
            }
            if ((temp == NULL) || ((i == pos - 1) && (temp->next == NULL)))
            {
                new_node->next = NULL;
                (*tail)->next = new_node;
                (*tail) = new_node;
                return (new_node);
            }
            else
            {
                new_node->next = temp->next;
                temp->next = new_node;
                return (new_node);
            }
        }
    }

}


int main()
{
    struct node* head, * tail, * p;
    int n, i, ele, pos;

    do {
        printf(" Enter the number of elements initially in linked list:   ");
        scanf("%d", &n);
        if (n < 0)
        {
            printf(" Enter the number greater than zero \n");
        }
    } while (n < 0);
    printf(" Enter the elements of linked list \n");
    for (i = 0; i < n; i++)
    {
        p = (struct node*)malloc(sizeof(struct node));
        printf("%d element : ", i + 1);
        scanf("%d", &p->data);
        if (head == NULL)
        {
            head = p;
            tail = p;
        }
        else
        {
            tail->next = p;
            tail = p;
        }
    }

    printList();

    printf(" Enter the value which you want to insert:  ");
    scanf("%d", &ele);

    printf(" Enter the position at which you want to insert:  ");
    scanf("%d", &pos);

    insertion(&head, &tail, ele, pos);
    return 0;
}

Answer 1

int main()
{
    struct node* head, * tail, * p;

这是你的问题。您在名为 head 和 tail 的 main 中声明了局部变量，它们在该范围内隐藏了全局变量。当您稍后在 main 中分配给 head 和 tail 时，您正在更改局部变量的值。全局变量保持不变。

链表中显示节点的代码不起作用

Code of display node in linked list doesn't work

c

linked-list

data-structures

singly-linked-list