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
#include
// 下面的宏来自于:
#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
struct list_head {
struct list_head *next;
struct list_head *prev;
};
// 下面的宏及函数摘自于
#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=0
name offset=4
ch offset=132
ma offset=136
en offset=140
list 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=77
id=2, name=stu02, ch=15, ma=93, en=35
id=3, name=stu03, ch=86, ma=92, en=49
id=4, name=stu04, ch=21, ma=62, en=27
id=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=63
id=4, name=stu04, ch=21, ma=62, en=27
id=3, name=stu03, ch=86, ma=92, en=49
id=2, name=stu02, ch=15, ma=93, en=35
id=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);
}