使用信号量的餐饮哲学家问题

先决条件——进程同步、信号量、使用监视器的餐饮哲学家解决方案
哲学家进餐问题——哲学家进餐问题指出 K 位哲学家围坐在一张圆桌旁，每对哲学家之间夹着一根筷子。每个哲学家之间有一根筷子。如果哲学家能拿起旁边的两根筷子，他就可以吃饭。一根筷子可以被任何一个相邻的追随者捡起，但不能同时被两个人捡起。

餐饮哲学家的信号量解决方案 –
每个哲学家都由以下伪代码表示：

process P[i]
 while true do
   {  THINK;
      PICKUP(CHOPSTICK[i], CHOPSTICK[i+1 mod 5]);
      EAT;
      PUTDOWN(CHOPSTICK[i], CHOPSTICK[i+1 mod 5])
   }

哲学家有三种状态：思考、饥饿和进食。这里有两个信号量：互斥量和哲学家的信号量数组。使用互斥量使得两个哲学家不能同时访问拾取或放下。该数组用于控制每个哲学家的行为。但是，由于编程错误，信号量会导致死锁。
代码 –

C

#include 
#include 
#include 
 
#define N 5
#define THINKING 2
#define HUNGRY 1
#define EATING 0
#define LEFT (phnum + 4) % N
#define RIGHT (phnum + 1) % N
 
int state[N];
int phil[N] = { 0, 1, 2, 3, 4 };
 
sem_t mutex;
sem_t S[N];
 
void test(int phnum)
{
    if (state[phnum] == HUNGRY
        && state[LEFT] != EATING
        && state[RIGHT] != EATING) {
        // state that eating
        state[phnum] = EATING;
 
        sleep(2);
 
        printf("Philosopher %d takes fork %d and %d\n",
                      phnum + 1, LEFT + 1, phnum + 1);
 
        printf("Philosopher %d is Eating\n", phnum + 1);
 
        // sem_post(&S[phnum]) has no effect
        // during takefork
        // used to wake up hungry philosophers
        // during putfork
        sem_post(&S[phnum]);
    }
}
 
// take up chopsticks
void take_fork(int phnum)
{
 
    sem_wait(&mutex);
 
    // state that hungry
    state[phnum] = HUNGRY;
 
    printf("Philosopher %d is Hungry\n", phnum + 1);
 
    // eat if neighbours are not eating
    test(phnum);
 
    sem_post(&mutex);
 
    // if unable to eat wait to be signalled
    sem_wait(&S[phnum]);
 
    sleep(1);
}
 
// put down chopsticks
void put_fork(int phnum)
{
 
    sem_wait(&mutex);
 
    // state that thinking
    state[phnum] = THINKING;
 
    printf("Philosopher %d putting fork %d and %d down\n",
           phnum + 1, LEFT + 1, phnum + 1);
    printf("Philosopher %d is thinking\n", phnum + 1);
 
    test(LEFT);
    test(RIGHT);
 
    sem_post(&mutex);
}
 
void* philospher(void* num)
{
 
    while (1) {
 
        int* i = num;
 
        sleep(1);
 
        take_fork(*i);
 
        sleep(0);
 
        put_fork(*i);
    }
}
 
int main()
{
 
    int i;
    pthread_t thread_id[N];
 
    // initialize the semaphores
    sem_init(&mutex, 0, 1);
 
    for (i = 0; i < N; i++)
 
        sem_init(&S[i], 0, 0);
 
    for (i = 0; i < N; i++) {
 
        // create philosopher processes
        pthread_create(&thread_id[i], NULL,
                       philospher, &phil[i]);
 
        printf("Philosopher %d is thinking\n", i + 1);
    }
 
    for (i = 0; i < N; i++)
 
        pthread_join(thread_id[i], NULL);
}

注 –以下程序只能使用带有信号量和 pthread 库的 C 编译器进行编译。
参考 –
餐饮哲学家的解决方案 – cs.gordon.edu
餐饮哲学家的解决方案 – cs.indiana.edu