16.3 In the famous dining philosophers problem, a bunch of philosophers are sitting around a circular table with one chopstick between each of them. A philosopher needs both chopsticks to eat, and always picks up the left chopstick before the right one. A deadlock could potentially occur if all the philosophers reached for the left chopstick at the same time. Using threads and locks, implement a simulation of the dining philosophers problem that prevents deadlocks.
经典的哲学家聚餐问题，说是有一堆哲学家围着一个圆桌子坐着吃饭，每两个人之间都有一根筷子，每个人吃饭需要都需要左右两边的筷子，而且是先拿起左边的筷子，再拿右边的筷子，那么如果当所有的哲学家都拿着左边的筷子，那么就会产生死锁的情况。如果我们先不考虑死锁的问题，先来实现这个问题。我们可以把每个哲学家都当做一个线程，然后筷子被哲学家拿起后可以调用锁，当被放下后调用解锁，参见代码如下：
import java.util.concurrent.locks.Lock; import java.util.concurrent.locks.ReentrantLock; public class Chopstick { private Lock lock; public Chopstick() { lock = new ReentrantLock(); } public void pickUp() { lock.lock(); } public void putDown() { lock.unlock(); } } public class Philosopher extends Thread { private final int maxPause = 100; private int bites = 10; private Chopstick left; private Chopstick right; private int index; public Philosopher(int i, Chopstick left, Chopstick right) { this.left = left; this.right = right; } public void eat() { System.out.println("Philosopher" + index + ": start eating"); pickUp(); chew(); putDown(); System.out.println("Philosopher " + index + ": done eating"); } public void pickUp() { pause(); left.pickUp(); pause(); right.pickUp(); } public void chew() { System.out.println("Philosopher " + index + ": eating"); pause(); } public void pause() { try { int pause = (int)(Math.random() * maxPause); Thread.sleep(pause); } catch (InterruptedException e) { e.printStackTrace(); } } public void putDown() { left.putDown(); right.putDown(); } public void run() { for (int i = 0; i < bites; ++i) { eat(); } } } public class j { public static int size = 3; public static int leftOf(int i) { return i; } public static int rightOf(int i) { return (i + 1) % size; } public static void main(String[] args) { Chopstick[] chopsticks = new Chopstick[size + 1]; for (int i = 0; i < size + 1; ++i) { chopsticks[i] = new Chopstick(); } Philosopher[] philosophers = new Philosopher[size]; for (int i = 0; i < size; ++i) { Chopstick left = chopsticks[leftOf(i)]; Chopstick right = chopsticks[rightOf(i)]; philosophers[i] = new Philosopher(i, left, right); } for (int i = 0; i < size; ++i) { philosophers[i].start(); } } }
上面的代码在执行中基本都会陷入死循环，因为发生了死锁的情况，所以我们应该想机制来避免死锁的发生，那么怎么做呢，我们首先想想死锁是怎么形成的，是因为每个人都拿着左边的筷子不放，又无法拿到右边的筷子，所以就一直僵持着，那么我们换个思路想想，如果每个人在拿了左筷子，发现没法取得右筷子后，就把左筷子放下，这样就可以避免死锁的形成。那么我们在Chopstik类中的pickUp函数中就应该使用tryLock()来代替lock，这样只有在有左筷子的时候才能锁上左筷子，而且在Philosopher类中的pickUp函数中，先判断能不能拿左筷子，不能拿直接返回false，能拿的话再来看能不能拿右筷子，不能拿的话，先把左筷子放下，再返回false，能拿的话返回true。这样在eat函数中先看pickUp是否能返回true，能返回的话再继续运行之后的东西，参见代码如下：
import java.util.concurrent.locks.Lock; import java.util.concurrent.locks.ReentrantLock; public class Chopstick { private Lock lock; public Chopstick() { lock = new ReentrantLock(); } public boolean pickUp() { return lock.tryLock(); } public void putDown() { lock.unlock(); } } public class Philosopher extends Thread { private final int maxPause = 100; private int bites = 10; private Chopstick left; private Chopstick right; private int index; public Philosopher(int i, Chopstick left, Chopstick right) { index = i; this.left = left; this.right = right; } public void eat() { System.out.println("Philosopher" + index + ": start eating"); if (pickUp()) { chew(); putDown(); System.out.println("Philosopher " + index + ": done eating"); } else { System.out.println("Philosopher " + index + ": gave up on eating"); } } public boolean pickUp() { pause(); if (!left.pickUp()) { return false; } pause(); if (!right.pickUp()) { left.putDown(); return false; } pause(); return true; } public void chew() { System.out.println("Philosopher " + index + ": eating"); pause(); } public void pause() { try { int pause = (int)(Math.random() * maxPause); Thread.sleep(pause); } catch (InterruptedException e) { e.printStackTrace(); } } public void putDown() { left.putDown(); right.putDown(); } public void run() { for (int i = 0; i < bites; ++i) { eat(); } } } public class j { public static int size = 3; public static int leftOf(int i) { return i; } public static int rightOf(int i) { return (i + 1) % size; } public static void main(String[] args) { Chopstick[] chopsticks = new Chopstick[size + 1]; for (int i = 0; i < size + 1; ++i) { chopsticks[i] = new Chopstick(); } Philosopher[] philosophers = new Philosopher[size]; for (int i = 0; i < size; ++i) { Chopstick left = chopsticks[leftOf(i)]; Chopstick right = chopsticks[rightOf(i)]; philosophers[i] = new Philosopher(i, left, right); } for (int i = 0; i < size; ++i) { philosophers[i].start(); } } }
本文转自博客园Grandyang的博客，原文链接：哲学家聚餐问题[CareerCup] 16.3 Dining Philosophers ，如需转载请自行联系原博主。
