Linux 多线程同步之哲学家用餐问题

问题描述：

有五个哲学家公用一张餐桌，分别坐在周围的五张椅子上，在餐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用餐。平时，一个哲学家进行思考，饥饿时便试图拿取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。（计算机操作系统 第三版）

书上为代码：

Var chopstick: array[0,...,4] of semaphore;
repeat
	wait(chopstick[i]);
	wait(chopstick[(i+1)%5])
			.
			.
	eat:
			.
			.
	signal(chopstick[i])
	signal(opstick[(i+1)%5])
			.
			.
	think:
until false;

这样当五个哲学家同时饥饿时而各自拿起左边的筷子是，就会使五个信号量chopstick均为0；当他们再试图去拿右边的筷子时，都将因没有筷子可拿而无限地等待。从而导致死锁。

本文解决死锁的策略是：只有当哲学家的左右两只筷子均可以用的时候，才允许其拿起筷子进餐，否则则该哲学家一直请求左边的筷子保持阻塞状态（左边的筷子信号量为1则直接占有，信号量为0则进入阻塞状态），右边的筷子若信号量为0，则非阻塞，进入等待。

详细代码：

/*
 * philosopher.c
 *
 *  Created on: 2012-12-17
 *      Author: xkey
 */

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <pthread.h>
#include <semaphore.h>
#include <string.h>

#define PEOPLE_NUM 5
#define THINKING 1
#define HUNGRY 2
#define EATING 3

sem_t chopsticks[PEOPLE_NUM];
pthread_mutex_t mutex;

void *philosopher(void *arg){
	int id = (int) arg;

	int state = THINKING;
	int right = (id + 1) % PEOPLE_NUM;
	int left = (id + PEOPLE_NUM - 1) % PEOPLE_NUM;
	char ptrState[32];

	while(1){
		switch(state){
		case THINKING:
			usleep(300);
			state = HUNGRY;
			strcpy(ptrState,"Thinking before eat");
			break;
		case HUNGRY:
			strcpy(ptrState,"Hungry");
			if(sem_wait(&chopsticks[left]) == 0){//阻塞状态
				if(sem_trywait(&chopsticks[right]) == 0){//非阻塞
					strcpy(ptrState,"I will Eating");
					state = EATING;
				}else{
					state = THINKING;
					strcpy(ptrState,"I have not chopsticks");
					sem_post(&chopsticks[left]);//释放请求的得到的left筷子
					printf("Philosopher right chopsticks is busy,right=%d,thread id is %d\n",right,id);
				}
			}else{
				printf("Philosopher left chopsticks is busy,left=%d,thread id is %d\n",left,id);//这句话由于上面被阻塞永远不会输出
			}
			break;
		case EATING:
			printf("Philosopher fetch left and right chopsticks: (%d,%d), threadid is %d\n",left,right,id);
			sem_post(&chopsticks[left]);
			sem_post(&chopsticks[right]);
			printf("Philosopher release left and right chopsticks: (%d,%d), threadid is %d\n",left,right,id);
			usleep(500);
			state = THINKING;
			strcpy(ptrState,"Thinking after eat");
			break;
		}
		pthread_mutex_lock(&mutex);
		printf("Philosopher is %s, thread id is %d\n",ptrState,id);
		pthread_mutex_unlock(&mutex);
		usleep(1000);
	}

	pthread_exit((void*)0);
}

int main(){
	pthread_t tid[PEOPLE_NUM];
	int i;
	pthread_mutex_init(&mutex,NULL);
	for(i = 0 ; i < PEOPLE_NUM ; i ++){
		sem_init(&chopsticks[i],0,1);
	}
	for(i = 0 ; i < PEOPLE_NUM ; i ++){
		pthread_create(&tid[i],NULL,philosopher,(void*)i);
	}
	for(i = 0 ; i < PEOPLE_NUM ; i ++){
		pthread_join(tid[i],NULL);
	}
	return 0;
}

 gcc编译时，加上-lphread选项，例：gcc philosopher.c -lpthread

这里有个问题，也就是解决死锁的策略

if(sem_wait(&chopsticks[left]) == 0){//阻塞状态
if(sem_trywait(&chopsticks[right]) == 0){//非阻塞

可以考虑如果第二个if使用sem_wait会怎么样，是不是还能保证解决死锁，个人认为均用sem_wait就不需要写if语句了，这样应该会造成死锁；

另：如果两个if均使用sem_trywait会怎么样？

个人觉得这两个问题是本文应该思考的地方，本身哲学家进餐问题的实现不难，关键就是解决死锁的策略。

本文解决哲学家用餐问题使用的是信号量方法，该方法详解的Bolg地址见http://blog.csdn.net/wtz1985/article/details/3826291点击打开链接

另附两篇博文都是关于解决哲学家进餐问题的：

1、使用互斥量：http://blog.csdn.net/kongzhp/article/details/7487150

2、使用信号量（本文参照的方法，但是我看他写的全部是用sem_wait的）：http://www.cnblogs.com/margincc/archive/2011/06/30/2094418.html

附：由于本人对Linux多线程研究不深，如有错误，欢迎指教，谢谢。

下面贴出上面代码的执行结果：

Philosopher is Thinking before eat, thread id is 4
Philosopher fetch left and right chopsticks: (0,2), threadid is 1
Philosopher release left and right chopsticks: (0,2), threadid is 1
Philosopher is Thinking after eat, thread id is 1
Philosopher right chopsticks is busy,right=0,thread id is 4
Philosopher is I have not chopsticks, thread id is 4
Philosopher is Thinking before eat, thread id is 1
Philosopher fetch left and right chopsticks: (4,1), threadid is 0
Philosopher release left and right chopsticks: (4,1), threadid is 0
Philosopher is Thinking before eat, thread id is 3
Philosopher is Thinking before eat, thread id is 4
Philosopher is I will Eating, thread id is 1
Philosopher is Thinking after eat, thread id is 0
Philosopher is Thinking before eat, thread id is 2
Philosopher right chopsticks is busy,right=0,thread id is 4
Philosopher is I have not chopsticks, thread id is 4
Philosopher fetch left and right chopsticks: (0,2), threadid is 1
Philosopher release left and right chopsticks: (0,2), threadid is 1
Philosopher is I will Eating, thread id is 3
Philosopher is I will Eating, thread id is 2
Philosopher is Thinking before eat, thread id is 0
Philosopher is Thinking after eat, thread id is 1
Philosopher is Thinking before eat, thread id is 4
Philosopher fetch left and right chopsticks: (2,4), threadid is 3
Philosopher release left and right chopsticks: (2,4), threadid is 3
Philosopher fetch left and right chopsticks: (1,3), threadid is 2
Philosopher release left and right chopsticks: (1,3), threadid is 2
Philosopher is I will Eating, thread id is 0
Philosopher is Thinking after eat, thread id is 3
Philosopher is Thinking before eat, thread id is 1
Philosopher is I will Eating, thread id is 4
Philosopher is Thinking after eat, thread id is 2
Philosopher fetch left and right chopsticks: (4,1), threadid is 0
Philosopher is Thinking before eat, thread id is 3
Philosopher fetch left and right chopsticks: (3,0), threadid is 4
Philosopher release left and right chopsticks: (3,0), threadid is 4
Philosopher release left and right chopsticks: (4,1), threadid is 0

可以看到，一开始线程4就被阻塞，原因就是右边的筷子0被线程1占有

红色部分，线程4一直不能同时得到左右两边的筷子

紫色部分，线程2左边的筷子1是空闲的，但是右边的筷子3之前已经被线程4所请求过，但是正是由于线程4不能得到其右边的筷子0，所以放弃了已经占有的其左边筷子3，所以线程2能够执行。

蓝色体现了多线程的同步执行，除此之外，发现线程4同时得到了左右两边的筷子3和0