单例模式完全解析

飞熊 — Wed, 23 Dec 2009 08:34:00 GMT

本文��探讨单例模式的各种情况�Q��ƈ�l�出相应的徏议。单例模式应该是设计模式中比较简单的一个，但是在多�U�程�q�发的环境下使用却是不那么简单了�?br />首先看最原始的单例模式�?br />

1 package xylz.study.singleton;
2
3 public class Singleton {
4
5     private static Singleton instance = null;
6
7     private Singleton() {
8     }
9
10     public static Singleton getInstance() {
11         if (instance == null) {
12             instance = new Singleton();
13         }
14         return instance;
15     }
16 }
17

昄��q�个写法在单�U�程环境下非常好�Q�但是多�U�程会导致多个实例出玎ͼ��q�个大家都能理解�?br />最��单的攚w��方式是��d��一个同步锁�?br />

1 package xylz.study.singleton;
2
3 public class SynchronizedSingleton {
4
5     private static SynchronizedSingleton instance = null;
6
7     private SynchronizedSingleton() {
8     }
9
10     public static synchronized SynchronizedSingleton getInstance() {
11         if (instance == null) {
12             instance = new SynchronizedSingleton();
13         }
14         return instance;
15     }
16 }
17

昄��上面的方法避免了�q�发的问题，但是�׃��我们只是在第一�ơ构造对象的时候才需要同步，以后��׃��再需要同步，所以这里不可避免的有性能开销。于是将锁去掉采用静态的属性来解决同步锁的问题�?br />

1 package xylz.study.singleton;
2
3 public class StaticSingleton {
4
5     private static StaticSingleton instance = new StaticSingleton();
6
7     private StaticSingleton() {
8     }
9
10     public static StaticSingleton getInstance() {
11         return instance;
12     }
13 }
14

上面的方法既没有锁又解决了性能问题�Q�看��h��已经满��需求了。但是追求“完��”的�E�序员想延时加蝲对象�Q�希望在�W�一�ơ获取的时候才构造对象，于是大家非常聪明的进行改造，也即非常出名的双重检查锁机制出来了�?br />

1 package xylz.study.singleton;
2
3 public class DoubleLockSingleton {
4
5     private static DoubleLockSingleton instance = null;
6
7     private DoubleLockSingleton() {
8     }
9
10     public static DoubleLockSingleton getInstance() {
11         if (instance == null) {
12             synchronized (DoubleLockSingleton.class) {
13                 if (instance == null) {
14                     instance = new DoubleLockSingleton();
15                 }
16             }
17         }
18         return instance;
19     }
20 }
21

双重锁机制看��h��非常巧妙的避免了上面的问题。但是真的是�q�样的吗�Q�文章�?a target="_blank">双重��查锁定及单例模式》中谈到了非常多演变的双重锁机制带来的问题，包括比较难以理解的指令重排序机制�{�。��M��是双重��查锁机制仍然对导致错误问题而不是性能问题�?br />于是�l�箋攚w��，某个牛�h利用JVM的特性来解决上述问题�Q�具体哪个牛人我忘记了，但是不是下面文章的作者�?br />�Q?�Q��?a target="_blank">Java theory and practice: Fixing the Java Memory Model, Part 2�?br />�Q?�Q��?a target="_blank">Initialize-On-Demand Holder Class and Singletons�?br />

1 package xylz.study.singleton;
2
3 public class HolderSingleton {
4
5     private static class HolderSingletonHolder {
6
7         static HolderSingleton instance = new HolderSingleton();
8     }
9
10     private HolderSingleton() {
11         //maybe throw an Exception when doing something
12     }
13
14     public static HolderSingleton getInstance() {
15         return HolderSingletonHolder.instance;
16     }
17 }
18

上述代码看�v来解决了上面单例模式遇到的所有问题，而且实际上工作的很好�Q�没有什么问题。但是却有一个致命的问题�Q�如果第11行抛��Z��一个异常，也就是第一�ơ构造函数失败将��D��永远无法再次得到构徏对象的机会�?br />使用下面的代码测试下�?br />

1 package xylz.study.singleton;
2
3 public class HolderSingletonTest {
4
5     private static class HolderSingletonHolder {
6
7         static HolderSingletonTest instance = new HolderSingletonTest();
8     }
9
10     private static boolean init = false;
11
12     private HolderSingletonTest() {
13         //maybe throw an Exception when doing something
14         if(!init) {
15             init=true;
16             throw new RuntimeException("fail");
17         }
18     }
19
20     public static HolderSingletonTest getInstance() {
21         return HolderSingletonHolder.instance;
22     }
23     public static void main(String[] args) {
24         for(int i=0;i<3;i++) {
25             try {
26                 System.out.println(HolderSingletonTest.getInstance());
27             } catch (Exception e) {
28                 System.err.println("one->"+i);
29                 e.printStackTrace();
30             }catch(ExceptionInInitializerError err) {
31                 System.err.println("two->"+i);
32                 err.printStackTrace();
33             }catch(Throwable t) {
34                 System.err.println("three->"+i);
35                 t.printStackTrace();
36             }
37         }
38     }
39 }
40

很不�q�将得到以下输出�Q?br />

1 two->0
2 java.lang.ExceptionInInitializerError
3     at xylz.study.singleton.HolderSingletonTest.getInstance(HolderSingletonTest.java:21)
4     at xylz.study.singleton.HolderSingletonTest.main(HolderSingletonTest.java:26)
5 Caused by: java.lang.RuntimeException: fail
6     at xylz.study.singleton.HolderSingletonTest.<init>(HolderSingletonTest.java:16)
7     at xylz.study.singleton.HolderSingletonTest.<init>(HolderSingletonTest.java:12)
8     at xylz.study.singleton.HolderSingletonTest$HolderSingletonHolder.<clinit>(HolderSingletonTest.java:7)
9      2 more
10 three->1
11 java.lang.NoClassDefFoundError: Could not initialize class xylz.study.singleton.HolderSingletonTest$HolderSingletonHolder
12     at xylz.study.singleton.HolderSingletonTest.getInstance(HolderSingletonTest.java:21)
13     at xylz.study.singleton.HolderSingletonTest.main(HolderSingletonTest.java:26)
14 three->2
15 java.lang.NoClassDefFoundError: Could not initialize class xylz.study.singleton.HolderSingletonTest$HolderSingletonHolder
16     at xylz.study.singleton.HolderSingletonTest.getInstance(HolderSingletonTest.java:21)
17     at xylz.study.singleton.HolderSingletonTest.main(HolderSingletonTest.java:26)
18

很显然我们想着�W�一�ơ加载失败第二次能够加蝲成功�Q�非�怸��q�，JVM一旦加载某个类��p�|��认为此�cȝ��定义有问题，��来不再加蝲�Q�这样就��D��我们没有��Z��再加载。目前看��h��没有办法避免此问题。如果要使用JVM的类加蝲�Ҏ��就必须保证�c�d��载一定正��，否则此问题将比�ƈ发和性能更严重。如果我们的�c�需要初始话那么��需要想其它办法避免在构造函��C��完成。看��h��像是又回��C��老地方，��N��不是么？

��M��Q�结论是目前没有一个十全十��的单例模式�Q�而大多数情况下我们只需要满��x��们的需求就行，没必有特意追求最“完��”解��x��案�?br />原文[http://www.imxylz.cn/p/177.html]

飞熊 2009-12-23 16:34 发表评论

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单例模式完全解析