《Cracking the Coding Interview》——第16章：线程与锁——题目3

2014-04-27 19:26

题目：哲学家吃饭问题，死锁问题经典模型（专门用来黑哲学家的？）。

解法：死锁四条件：1. 资源互斥。2. 请求保持。3. 非抢占。4. 循环等待。所以，某砖家拿起一只筷子后如果发现没有另一只了，就必须把手里这只筷子放下，这应该是通过破坏“请求保持”原则来防止死锁产生，请求资源失败时，连自己的资源也进一步释放，然后在下一轮里继续请求，直到成功执行。

代码：

 // This is the class for chopsticks.
 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();
     }
 }

 //------------------------------------I'm a delimiter------------------------------------
 // This is the class for philosophers.
 import java.util.Vector;

 public class Philosopher extends Thread {
     private Chopstick left;
     private Chopstick right;
     private int id;
     int appetite;

     final int FULL_APPETITE = 10;

     public Philosopher(Chopstick left, Chopstick right, int id) {
         // TODO Auto-generated constructor stub
         appetite = 0;
         this.left = left;
         this.right = right;
         this.id = id;
     }

     private boolean pickUp() {
         if (!left.pickUp()) {
             return false;
         }
         if (!right.pickUp()) {
             left.putDown();
             return false;
         }
         return true;
     }

     private void putDown() {
         left.putDown();
         right.putDown();
     }

     public boolean eat() {
         while (appetite < FULL_APPETITE) {
             if (!pickUp()) {
                 return false;
             }
             System.out.println(id + ":chew~");
             ++appetite;
             putDown();
         }
         return appetite == FULL_APPETITE;
     }

     @Override
     public void run() {
         // TODO Auto-generated method stub
         super.run();
         while (!eat()) {
             // Not full yet.
         }
     }

     public static void main(String[] args) {
         final int n = 6;
         Vector<Chopstick> chopsticks = new Vector<Chopstick>();
         Vector<Philosopher> philosophers = new Vector<Philosopher>();

         for (int i = 0; i < n; ++i) {
             chopsticks.add(new Chopstick());
         }
         for (int i = 0; i < n; ++i) {
             philosophers.add(new Philosopher(chopsticks.elementAt(i),
                     chopsticks.elementAt((i + 1) % n), i + 1));
         }

         for (int i = 0; i < n; ++i) {
             philosophers.elementAt(i).start();
         }
     }
 }