[Unix系统编程]用信号量实现哲学家就餐问题

概述：不知道是信号量简单，还是信号量太简单，《Advanced Programming in the UNIX Environment》居然不讲。有几个与信号量相关的系统调用，sem_init, sem_wait, sem_trywait, sem_post, sem_getvalue, sem_destory，可以通过man sem_init的方式查看帮助文档，确实比较简单，就不再赘述。

下面的例子也很简单，一看就懂，只为练习信号量的使用。

#include 
#include 
#include "apue.h"
#include "my_err.h" 

#define N 5 // No. of philosopher
#define M 5 // times of eating
sem_t forks[N];

void * thr_philosopher( void *arg);
int main(int argc, char* argv[])
{
int i = 0; 
int err;
pthread_t tid[N];
void *tret;
//initilize semaphore
for (i = 0; i {
if(sem_init(&forks[i], 0, 1) != 0)
{
err_quit("init forks error");
}
}
//create thread
for (i = 0; i {
err = pthread_create(&tid[i], NULL, thr_philosopher, (void *)i);
if (err != 0)
{
err_quit("can't create thread %d: %s\n", i + 1, strerror(err));
}
}

//get the return value
for (i = 0; i {
err = pthread_join(tid[i], &tret);
if (err != 0)
{
err_quit("can't join with philosopher %d : %s\n", i + 1,
strerror(err));
}
printf("-------------------philosopher %d has done-------------------\n", (int)tret);
}

// delete the source of semaphore
for (i = 0; i {
err = sem_destroy(&forks[i]);
if (err != 0)
{
err_sys("can't destory semaphore");
}
}
exit(0);
}

void * thr_philosopher( void *arg)
{

/*
 * here cann't judge arg == NULL
 * because (void *)0 will lead to arg = NULL
 */
int n = M;
int i = 0;
i = (int)arg;
while ( n-- )
{
sleep(1);
if ( i == N - 1)
{
sem_wait(&forks[0]);
sem_wait(&forks[i]);
}
else
{
sem_wait(&forks[i]);
sem_wait(&forks[i + 1]);
}
printf("philosopher %d is eating\n", i + 1);
if ( i == N - 1)
{
sem_post(&forks[0]);
sem_post(&forks[i]);
}
else
{
sem_post(&forks[i]);
sem_post(&forks[i + 1]);
}

}

return ((void*)i);
}