---

Semaphore bowl = m;

Semaphore mutex[n] = 1; //任何一个筷子都是互斥量

Semaphore full = n - 1;//允许最多n-1个哲学家一起行动

P(int i)

{

P(full); P(bowl);P(mutex[i]);P(mutex[(i+1)%n]);

//dine

V(mutex

}