C 当链表为空时删除链表的节点

Question

我正在执行c中的管理链接列表的函数，但我在删除函数时遇到了问题。这只需要一个指向链表的指针和一个值，并删除具有该值的第一个节点。问题是，如果我传递一个空列表，就会出现段错误。有什么想法吗？

sll_node *sll_remove(sll_node *list, int search_value)
{
    sll_node* head = list;
    sll_node* delete = list->next;


    if(head == NULL)
    {
        return list;
    }

    if(head != NULL && head->value == search_value)
    {
        list = head->next;
        free(head);
        return list;
    }

    while(delete)
    {
        if(search_value == delete->value)
        {
            head->next = delete->next;
            free(delete);
            return list;
        }
        head = head->next;
        delete = delete->next;
    }

    return list;
}

Answer 1

在解除引用之前，您需要检查 list 是否为空

 sll_node *sll_remove(sll_node *list, int search_value)
    {
    if(!list){
          return list;
    }
        sll_node* head = list;
        sll_node* delete = list->next;


        if(!head)
        {
            return list;
        }

        if(head && head->value == search_value)
        {
            list = head->next;
            free(head);
            return list;
        }

        while(delete)
        {
            if(search_value == delete->value)
            {
                head->next = delete->next;
                free(delete);
                return list;
            }
            head = head->next;
            delete = delete->next;
        }

        return list;
    }