《Cracking the Coding Interview》——第16章:线程与锁——题目3
时间:2014-04-27 21:30:09
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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:
1 // This is the class for chopsticks. 2 import java.util.concurrent.locks.Lock; 3 import java.util.concurrent.locks.ReentrantLock; 4 5 public class Chopstick { 6 private Lock lock; 7 8 public Chopstick() { 9 lock = new ReentrantLock(); 10 } 11 12 public boolean pickUp() { 13 return lock.tryLock(); 14 } 15 16 public void putDown() { 17 lock.unlock(); 18 } 19 } 20 21 //------------------------------------I‘m a delimiter------------------------------------ 22 // This is the class for philosophers. 23 import java.util.Vector; 24 25 public class Philosopher extends Thread { 26 private Chopstick left; 27 private Chopstick right; 28 private int id; 29 int appetite; 30 31 final int FULL_APPETITE = 10; 32 33 public Philosopher(Chopstick left, Chopstick right, int id) { 34 // TODO Auto-generated constructor stub 35 appetite = 0; 36 this.left = left; 37 this.right = right; 38 this.id = id; 39 } 40 41 private boolean pickUp() { 42 if (!left.pickUp()) { 43 return false; 44 } 45 if (!right.pickUp()) { 46 left.putDown(); 47 return false; 48 } 49 return true; 50 } 51 52 private void putDown() { 53 left.putDown(); 54 right.putDown(); 55 } 56 57 public boolean eat() { 58 while (appetite < FULL_APPETITE) { 59 if (!pickUp()) { 60 return false; 61 } 62 System.out.println(id + ":chew~"); 63 ++appetite; 64 putDown(); 65 } 66 return appetite == FULL_APPETITE; 67 } 68 69 @Override 70 public void run() { 71 // TODO Auto-generated method stub 72 super.run(); 73 while (!eat()) { 74 // Not full yet. 75 } 76 } 77 78 public static void main(String[] args) { 79 final int n = 6; 80 Vector<Chopstick> chopsticks = new Vector<Chopstick>(); 81 Vector<Philosopher> philosophers = new Vector<Philosopher>(); 82 83 for (int i = 0; i < n; ++i) { 84 chopsticks.add(new Chopstick()); 85 } 86 for (int i = 0; i < n; ++i) { 87 philosophers.add(new Philosopher(chopsticks.elementAt(i), 88 chopsticks.elementAt((i + 1) % n), i + 1)); 89 } 90 91 for (int i = 0; i < n; ++i) { 92 philosophers.elementAt(i).start(); 93 } 94 } 95 }
《Cracking the Coding Interview》——第16章:线程与锁——题目3,码迷,mamicode.com
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