使用信号量的问题

先决条件—— 进程同步 , 信号灯 , 使用监视器的解决方案 用餐哲学家问题—— 用餐哲学家问题表明,K位哲学家围坐在一张圆桌旁,每对哲学家之间各有一根筷子。每个哲学家之间都有一根筷子。哲学家可以吃东西,只要他能拿起旁边的两根筷子。一根筷子可以被任何一个相邻的跟随者捡起,但不能同时被两个跟随者捡起。

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图片[1]-使用信号量的问题-yiteyi-C++库

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、饥饿和饮食 这里有两个信号量:互斥和哲学家的信号量数组。使用互斥锁时,两位哲学家不能同时访问拾取或放下。数组用于控制每个哲学家的行为。但是,由于编程错误,信号量可能会导致死锁。 代码–

C

#include <pthread.h>
#include <semaphore.h>
#include <stdio.h>
#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" ,
phnum + 1, LEFT + 1, phnum + 1);
printf ( "Philosopher %d is Eating" , 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" , 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" ,
phnum + 1, LEFT + 1, phnum + 1);
printf ( "Philosopher %d is thinking" , phnum + 1);
test(LEFT);
test(RIGHT);
sem_post(&mutex);
}
void * philosopher( 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,
philosopher, &phil[i]);
printf ( "Philosopher %d is thinking" , i + 1);
}
for (i = 0; i < N; i++)
pthread_join(thread_id[i], NULL);
}


注—— 下面的程序只能用带有信号量和pthread库的C编译器编译。 参考资料- 餐饮哲学家的解决方案——cs。戈登。埃杜 餐饮哲学家的解决方案——cs。印第安纳州。埃杜

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