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一、问题介绍
由Dijkstra提出并解决的哲学家进餐问题(The Dinning Philosophers Problem)是典型的同步问题。该问题是描述有五个哲学家共用一张圆桌,分别坐在周围的五张椅子上,在圆桌上有五个碗和五只筷子,他们的生活方式是交替地进行思考和进餐。平时,一个哲学家进行思考,饥饿时便试图取用其左右最靠近他的筷子,只有在他拿到两只筷子时才能进餐。进餐完毕,放下筷子继续思考。
二、POSIX中的互斥量
1、库文件:#include <pthread.h>
2、数据类型:
pthread_mutex_t //互斥量
pthread_mutexattr_t //互斥量的属性
3、互斥量相关的函数:
//…
pthread_mutex_tmutex;
pthread_mutex_init(&mutex, NULL);
pthread_mutex_lock(&mutex);
//do something
pthread_mutex_unlock(&mutex);
pthread_mutex_destroy(&mutex);
//…
三、POSIX线程函数
1、库文件:#include<pthread.h>
2、数据类型:pthread_t;线程ID
3、线程相关函数:
线程退出有以下几种情况:
1、库文件:#include<semaphore.h>
2、信号量数据类型:sem_t
sem_tsem;
sem_init(&sem,0, 1);//初始化一个值为1的信号量
sem_wait(&sem);//获取信号量
//dosomthing
sem_post(&sem);//释放信号量
sem_destroy(&sem);//销毁一个无名信号量
五、流程图
六、代码示例
使用互斥量:
#include
#include
#include
#include
#include
#include
//筷子作为mutex
pthread_mutex_t chopstick[6] ;
void *eat_think(void *arg)
{
char phi = *(char *)arg;
int left,right; //左右筷子的编号
switch (phi){
case 'A':
left = 5;
right = 1;
break;
case 'B':
left = 1;
right = 2;
break;
case 'C':
left = 2;
right = 3;
break;
case 'D':
left = 3;
right = 4;
break;
case 'E':
left = 4;
right = 5;
break;
}
int i;
for(;;){
usleep(3); //思考
pthread_mutex_lock(&chopstick[left]); //拿起左手的筷子
printf("Philosopher %c fetches chopstick %dn", phi, left);
if (pthread_mutex_trylock(&chopstick[right]) == EBUSY){ //拿起右手的筷子
pthread_mutex_unlock(&chopstick[left]); //如果右边筷子被拿走放下左手的筷子
continue;
}
// pthread_mutex_lock(&chopstick[right]); //拿起右手的筷子,如果想观察死锁,把上一句if注释掉,再把这一句的注释去掉
printf("Philosopher %c fetches chopstick %dn", phi, right);
printf("Philosopher %c is eating.n",phi);
usleep(3); //吃饭
pthread_mutex_unlock(&chopstick[left]); //放下左手的筷子
printf("Philosopher %c release chopstick %dn", phi, left);
pthread_mutex_unlock(&chopstick[right]); //放下左手的筷子
printf("Philosopher %c release chopstick %dn", phi, right);
pthread_mutex_destroy(&chopstick[left]);
pthread_mutex_destroy(&chopstick[right]);
}
}
int main(){
pthread_mutex_t A,B,C,D,E; //5个哲学家
int i;
for (i = 0; i < 5; i++)
pthread_mutex_init(&chopstick[i],NULL);
pthread_create(&A,NULL, eat_think, "A");
pthread_create(&B,NULL, eat_think, "B");
pthread_create(&C,NULL, eat_think, "C");
pthread_create(&D,NULL, eat_think, "D");
pthread_create(&E,NULL, eat_think, "E");
pthread_join(A,NULL);
pthread_join(B,NULL);
pthread_join(C,NULL);
pthread_join(D,NULL);
pthread_join(E,NULL);
return 0;
}
#include
#include
#include
#include
sem_t chopstick[6] ;
void *eat_think(void *arg)
{
char phi = *(char *)arg;
int left,right;
switch (phi){
case 'A':
left = 5;
right = 1;
break;
case 'B':
left = 1;
right = 2;
break;
case 'C':
left = 2;
right = 3;
break;
case 'D':
left = 3;
right = 4;
break;
case 'E':
left = 4;
right = 5;
break;
}
int i;
for(;;){
usleep(3);
sem_wait(&chopstick[left]);
printf("Philosopher %c fetches chopstick %dn", phi, left);
if (sem_trywait(&chopstick[right]) < 0){
sem_post(&chopstick[left]);
continue;
}
printf("Philosopher %c fetches chopstick %dn", phi, right);
printf("Philosopher %c is eating.n",phi);
usleep(3);
sem_post(&chopstick[left]);
printf("Philosopher %c release chopstick %dn", phi, left);
sem_post(&chopstick[right]);
printf("Philosopher %c release chopstick %dn", phi, right);
}
}
int main(){
pthread_t A,B,C,D,E;
int i;
for (i = 0; i < 5; i++)
sem_init(&chopstick[i],0,1);
pthread_create(&A,NULL, eat_think, "A");
pthread_create(&B,NULL, eat_think, "B");
pthread_create(&C,NULL, eat_think, "C");
pthread_create(&D,NULL, eat_think, "D");
pthread_create(&E,NULL, eat_think, "E");
pthread_join(A,NULL);
pthread_join(B,NULL);
pthread_join(C,NULL);
pthread_join(D,NULL);
pthread_join(E,NULL);
return 0;
}
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