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
时间： 2025-01-14 09:17:10