我找不到链表的错误（为什么我的头指针在移动？）

Question

我已经尝试了很多次将头指针指向第一个节点。首先（在空列表中）它正确地指向第一个节点。但是在第一次循环之后，head指针指向链接的newnode。其实现在我对我的整个代码也很不确定。

int main(void){
struct library *head = NULL; //set the head pointer to NULL
int option;
printf("Enter the number:");


while((option = getchar())!= 9){
switch(option){
case '1':
    {
    char title[1000];
    char author[1000];
    char subject[1000];
    printf("Enter title of the book you want to add:");
    scanf("%s",title);
    printf("Enter author of the book you want to add:");
    scanf("%s",author);
    printf("Enter subject of the book you want to add:");
    scanf("%s",subject);
    add_book(title,author,subject,&head);
    printf("successful! and head pointer is pointing to %s\n",head->collection.title);
    break;
    }
  }
}


void add_book(char title[],char author[],char subject[], struct library ** head){
struct library *current;
struct library *newnode = malloc(sizeof(struct library));
newnode->collection.title = title;
newnode->collection.author = author;
newnode->collection.subject = subject;      // assigning value inside newnode
newnode->num_books = 0;
newnode->next = NULL;                       // assign NULL value to the end of newnod

//when the head is NULL which means when the list is empty
if(*head == NULL)
{
    current = newnode;
    *head = current;

    return;
}

else
{
    current = *head;                //assign the first node to current pointer
    //find the last node of the list
    while(current->next != NULL)
    {
        current = current->next;
    }
    current->next = newnode;                    // link the last node to new node
    return;
}
}

这是为此的结构

struct book {
char* title;
char* author;
char* subject;
};

struct library {
struct book collection;
int num_books;
struct library* next;
};

Answer 1

lifetime of char title[1000];, char author[1000];, and char subject[1000]; ends when execution reaches the end of the block inside case '1': { /* ... */ }. Once this happens, the pointers that were assigned in add_book become dangling pointers - 指向无效内存。

要解决此问题，您必须确保字符串的生命周期与包含它们的结构的生命周期相匹配。这可以通过在结构本身

中分配足够的 space 来完成

struct book {
    char title[1000];
    /* etc. */
};

或者通过动态分配足够的 space 来复制每个字符串。在任何情况下，您都必须将字符串复制到此内存 (man 3 strcpy)。

如果它在您的系统上可用，man 3 strdup 会同时执行第二种形式的两个步骤。否则，与strcpy(malloc(strlen(source_string) + 1), source_string);.

大致相同

另请注意，scanf 说明符 %s 在不使用 字段宽度说明符 时与 dangerous as gets 相同（例如，char buffer[1000]; scanf("%999s", buffer);), 因为它可能会溢出你的缓冲区。

一个示例程序。一个一个输入字符串，以EOF CTRL+D (Windows: CTRL+Z, RETURN).

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

struct link {
    char *string;
    /* alternatively: char string[512]; */
    struct link *next;
};

void add_link(struct link **root, const char *string) {
    struct link *node = calloc(1, sizeof *node);
    node->string = strdup(string);
    /* alternatively: strcpy(node->string, string) */

    if (*root) {
        struct link *tail = *root;

        while (tail->next)
            tail = tail->next;

        tail->next = node;
    } else
        *root = node;
}

int main(void) {
    struct link *head = NULL;

    while (1) {
        char buffer[512];

        if (!fgets(buffer, sizeof buffer, stdin))
            break;

        /* remove newline */
        buffer[strcspn(buffer, "\n")] = '[=11=]';

        add_link(&head, buffer);
    }

    for (struct link *node = head, *next; node; node = next) {
        next = node->next;
        printf("STRING: %s\n", node->string);
        free(node->string);
        free(node);
    }
}

注意：在实际程序中，您应该始终检查内存分配函数（malloc、calloc、strdup 等）的 return 值，因为他们可能会失败。

我找不到链表的错误（为什么我的头指针在移动？）

I cant find the error of my linked list( why is my head pointer moving?)

c

linked-list