2014-04-27 20:16
题目：假设一个类Foo有三个公有的成员方法first()、second()、third()。请用锁的方法来控制调用行为，使得他们的执行循序总是遵从first、second、third的顺序。
解法：你应该想到了用lock的方法类阻塞，不过这里面有个概念问题使得直接用ReentrantLock不能通过编译（对于一个锁对象，不同在A线程中锁定，又在B线程中解锁，不允许这样的归属关系），可以用Semaphore来达到相同的目的。请看下面的代码。
代码：
1 // 16.5 There‘re three methods in a class FooBar, how would you make sure that they‘re executed in a fixed order, in whichever order they‘re called?
2 public class FirstRun implements Runnable {
3 private FooBar fooBar;
4
5 public FirstRun(FooBar fooBar) {
6 // TODO Auto-generated constructor stub
7 this.fooBar = fooBar;
8 }
9
10 @Override
11 public void run() {
12 // TODO Auto-generated method stub
13 fooBar.first();
14 }
15 }
16
17 // -----------------------------------------------------------------------------
18 public class SecondRun implements Runnable {
19 private FooBar fooBar;
20
21 public SecondRun(FooBar fooBar) {
22 // TODO Auto-generated constructor stub
23 this.fooBar = fooBar;
24 }
25
26 @Override
27 public void run() {
28 // TODO Auto-generated method stub
29 fooBar.second();
30 }
31 }
32
33 // -----------------------------------------------------------------------------
34 public class ThirdRun implements Runnable {
35 private FooBar fooBar;
36
37 public ThirdRun(FooBar fooBar) {
38 // TODO Auto-generated constructor stub
39 this.fooBar = fooBar;
40 }
41
42 @Override
43 public void run() {
44 // TODO Auto-generated method stub
45 fooBar.third();
46 }
47 }
48
49 // -----------------------------------------------------------------------------
50 import java.util.concurrent.Semaphore;
51
52 public class FooBar {
53 private Semaphore sem1;
54 private Semaphore sem2;
55 private Semaphore sem3;
56
57 public FooBar() {
58 // TODO Auto-generated constructor stub
59 sem1 = new Semaphore(1);
60 sem2 = new Semaphore(1);
61 sem3 = new Semaphore(1);
62
63 try {
64 sem1.acquire();
65 sem2.acquire();
66 sem3.acquire();
67 } catch (InterruptedException e) {
68 // TODO Auto-generated catch block
69 e.printStackTrace();
70 }
71 }
72
73 public void first() {
74 System.out.println("first");
75
76 sem1.release();
77 }
78
79 public void second() {
80 try {
81 sem1.acquire();
82 } catch (InterruptedException e) {
83 // TODO Auto-generated catch block
84 e.printStackTrace();
85 }
86 sem1.release();
87 System.out.println("second");
88 sem2.release();
89 }
90
91 public void third() {
92 try {
93 sem2.acquire();
94 } catch (InterruptedException e) {
95 // TODO Auto-generated catch block
96 e.printStackTrace();
97 }
98 sem2.release();
99 System.out.println("third");
100 sem3.release();
101 }
102
103 public static void main(String[] args) {
104 FooBar fooBar = new FooBar();
105 Thread t1 = new Thread(new FirstRun(fooBar));
106 Thread t2 = new Thread(new SecondRun(fooBar));
107 Thread t3 = new Thread(new ThirdRun(fooBar));
108
109 t3.start();
110 t1.start();
111 t2.start();
112 }
113 }
《Cracking the Coding Interview》——第16章：线程与锁——题目5,布布扣,bubuko.com
时间： 2024-11-03 05:33:01