C语言链表删除头节点Bug求助：删除后主指针值异常

我写了个简单链表，遇到个奇怪bug：先往头部加值1，再加值2，接着删除值为2的头节点。delete(head, 2)函数快返回时head->data == 1，但main里调用完这个函数后，head->data变成0了，输出结果完全不符合预期。

完整代码如下：

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

typedef struct linked_list
{
    int data;
    struct linked_list *next;
} linked_list_t;

void print_list(linked_list_t *head)
{
    linked_list_t *curr = head;
    printf("LIST: ");
    while (curr)
    {
        printf("%d ", curr->data);
        curr = curr->next;
    }
    printf("\n");
    fflush(stdout);
    
}

linked_list_t *add_to_front(linked_list_t *head, int data) 
{
    linked_list_t *new = malloc(sizeof(linked_list_t));
    if (!new) exit(1);
    new->next = head;
    new->data = data;
    return new;
}

void delete(linked_list_t *head, int data) 
{
    linked_list_t *current = head;
    linked_list_t *previous = NULL;
    // move current to the position of the node to delete
    while (current != NULL && current->data != data) 
    {
        previous = current;
        current = current->next;
    }

    if (current)
    {
        if (previous) 
        {
            // if the node to delete is not head
            previous->next = current->next;
        }
        else
        {
            // if the node to delete is head
            head = current->next;
        }
    }
    free(current);
}

int main()
{
    linked_list_t *head = NULL;
    print_list(head);
    // prints "LIST:"
    head = add_to_front(head, 1);
    head = add_to_front(head, 2);
    print_list(head);
    // prints "LIST: 2 1"
    // head->data = 1 just before delete() exits
    delete(head, 2);
    // head->data = 0 right after delete() exits
    print_list(head);
    // prints "LIST: 0 1" instead of "LIST: 1" 
}

问题根源 #

核心问题是delete函数里的head是值传递——你在函数内部修改head = current->next，只是修改了函数内部的局部变量，完全不会影响main函数里的head指针。当你删除头节点并free掉它之后，main里的head仍然指向这块已经被释放的内存，此时访问head->data属于访问野指针指向的已释放内存，值变成0是未定义行为的随机表现。

修复方案 #

有两种常用的修复方式：

方式1：让delete函数返回新的head指针 #

和你写的add_to_front逻辑保持一致，修改后返回新的头指针，在main里接收更新：

linked_list_t *delete(linked_list_t *head, int data) 
{
    linked_list_t *current = head;
    linked_list_t *previous = NULL;
    while (current != NULL && current->data != data) 
    {
        previous = current;
        current = current->next;
    }

    if (current)
    {
        if (previous) 
        {
            previous->next = current->next;
        }
        else
        {
            // 修改头节点时，更新局部head为新的头指针
            head = current->next;
        }
        free(current);
    }
    // 返回更新后的头指针
    return head;
}

main函数里调用时要更新head：

head = delete(head, 2);

方式2：使用二级指针直接修改原指针 #

通过传递指针的指针，直接在函数内部修改main里的head变量：

void delete(linked_list_t **head, int data) 
{
    linked_list_t *current = *head;
    linked_list_t *previous = NULL;
    while (current != NULL && current->data != data) 
    {
        previous = current;
        current = current->next;
    }

    if (current)
    {
        if (previous) 
        {
            previous->next = current->next;
        }
        else
        {
            // 通过二级指针修改原head指针
            *head = current->next;
        }
        free(current);
    }
}

main函数里调用时传入head的地址：

delete(&head, 2);

两种方式都能解决问题，第一种逻辑更贴合你现有代码的写法，第二种不需要返回值，直接修改原变量。另外要记住：访问已释放的内存是C语言中典型的未定义行为，结果可能随机变化，一定要避免这种情况。

内容的提问来源于stack exchange，提问作者Sgg8