本文主要是介绍linux内核中的offsetof、container_of、双链表list.h实践，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
先直接上程序，代码中包含了注释已经说清楚。在linux的应用层中编译、测试：
感谢李慧芹的B站课程：史上最强最细腻的linux嵌入式C语言学习教程【李慧芹老师】_哔哩哔哩_bilibili
#include <stdio.h>
#include <stdlib.h>// 下面的宏来自于: <linux/kernel.h>
#define offsetof(TYPE, MEMBER) ((size_t) &((TYPE *)0)->MEMBER)#define container_of(ptr, type, member) ({ \const typeof( ((type *)0)->member ) *__mptr = (ptr); \(type *)( (char *)__mptr - offsetof(type,member) );})// 下面的结构体定义来自于 <linux/types.h>
struct list_head {struct list_head *next;struct list_head *prev;
};// 下面的宏及函数摘自于 <linux/list.h>
#define list_entry(ptr, type, member) \container_of(ptr, type, member)#define list_for_each(pos, head) \for (pos = (head)->next; pos != (head); pos = pos->next) #define LIST_HEAD_INIT(name) { &(name), &(name) }#define LIST_HEAD(name) \struct list_head name = LIST_HEAD_INIT(name)void INIT_LIST_HEAD(struct list_head *list)
{list->next = list;list->prev = list;
}// 将 new 插入到 prev 和 next 的中间
void __list_add(struct list_head *new,struct list_head *prev,struct list_head *next)
{next->prev = new;new->next = next;new->prev = prev;prev->next = new;
}// 将 new 插入到 head 的后面
void list_add(struct list_head *new, struct list_head *head)
{__list_add(new, head, head->next);
}// 将 new 插入到 head 的前面
void list_add_tail(struct list_head *new, struct list_head *head)
{__list_add(new, head->prev, head);
}void __list_del(struct list_head * prev, struct list_head * next)
{next->prev = prev;prev->next = next;
}void __list_del_entry(struct list_head *entry)
{__list_del(entry->prev, entry->next);
}//
// 下面是业务层应用代码
//
// 应用层业务的结构体定义：
struct student
{int id;char name[128];int ch; //语文分数int ma; //数学分数int en; //英语分数struct list_head list; //包含一个 list_head
};void print_stu(struct student *st);int main()
{//// 0.1 测试宏 offsetof 使用//printf("id offset=%ld\n", offsetof(struct student, id));printf("name offset=%ld\n", offsetof(struct student, name));printf("ch offset=%ld\n", offsetof(struct student, ch));printf("ma offset=%ld\n", offsetof(struct student, ma));printf("en offset=%ld\n", offsetof(struct student, en));printf("list offset=%ld\n", offsetof(struct student, list));
/* id offset=0name offset=4ch offset=132ma offset=136en offset=140list offset=144上面看出，宏offsetof(TYPE, MEMBER)，就是返回成员MEMBER相对首的偏移！
*/// 0.2 测试宏 container_of 使用struct student stu={100, "std100", 78, 88, 98, NULL,};struct student *p=container_of(&stu.list, struct student, list);printf("&stu=%p\n", &stu);printf("&stu.list=%p\n", &stu.list);printf("&p=%p\n", p);
/*&stu= 0x7fffa2f4e1b0&stu.list= 0x7fffa2f4e240 0x240-0x1b0=144 即是上述list的偏移off&p= 0x7fffa2f4e1b0 上面看出，宏container_of(ptr, type, member) 即是返回结构体变量的首地址。那么问题来了，为何搞这么复杂的一个转换来获取首地址呢？直接使用&stu不就得到完了嘛！别急，看下面的应用！
*/int i=0;LIST_HEAD(head);//// 1. 创建5个结构体，使用 list 连起来//for(i=0; i<5; i++){struct student *st=malloc(sizeof(struct student));sprintf(st->name, "stu%02d", i+1);st->id=i+1;st->ch=rand()%100;st->ma=rand()%100;st->en=rand()%100;printf("id=%d, name=%s, ch=%d, ma=%d, en=%d\n",st->id, st->name, st->ch, st->ma, st->en);list_add(&(st->list), &head); //这里每次插入到head的后面！}
/*id=1, name=stu01, ch=83, ma=86, en=77id=2, name=stu02, ch=15, ma=93, en=35id=3, name=stu03, ch=86, ma=92, en=49id=4, name=stu04, ch=21, ma=62, en=27id=5, name=stu05, ch=90, ma=59, en=63注意上述创建的原始顺序！
*/printf("\n"); //// 2. 遍历打印//struct list_head *c;list_for_each(c, &head){struct student *st=container_of(c, struct student, list);print_stu(st);}
/*id=5, name=stu05, ch=90, ma=59, en=63id=4, name=stu04, ch=21, ma=62, en=27id=3, name=stu03, ch=86, ma=92, en=49id=2, name=stu02, ch=15, ma=93, en=35id=1, name=stu01, ch=83, ma=86, en=77因为是每次插入到head的后面，所以链表里面的顺序是5、4、3....
*/ //// 3. 查找一个节点//list_for_each(c, &head){struct student *st=container_of(c, struct student, list);if(st->id==3){printf("\nfind it!\n");print_stu(st);}}//// 4. 删除一个节点//list_for_each(c, &head){struct student *st=container_of(c, struct student, list);if(st->id==3){__list_del_entry(&st->list);free(st);}}//// 5. 再次输出打印//printf("\nreprintf:\n"); list_for_each(c, &head){struct student *st=container_of(c, struct student, list);print_stu(st);}
}void print_stu(struct student *st)
{printf("id=%d, name=%s, ch=%d, ma=%d, en=%d\n",st->id, st->name, st->ch, st->ma, st->en);
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