java 5 后的并发锁 读写锁 1（从 coolxing 转载 http://coolxing.iteye.com/blog/1236696）

问题的描述
启动3个线程打印递增的数字, 线程1先打印1,2,3,4,5, 然后是线程2打印6,7,8,9,10, 然后是线程3打印11,12,13,14,15. 接着再由线程1打印16,17,18,19,20....以此类推, 直到打印到75. 程序的输出结果应该为:

线程1: 1
线程1: 2
线程1: 3
线程1: 4
线程1: 5

线程2: 6
线程2: 7
线程2: 8
线程2: 9
线程2: 10
...

线程3: 71
线程3: 72
线程3: 73
线程3: 74
线程3: 75


解法一: 采用原始的synchronized, wait(), notify(), notifyAll()等方式控制线程.

Java代码 
public class NumberPrintDemo { 
    // n为即将打印的数字 
    private static int n = 1; 
    // state=1表示将由线程1打印数字, state=2表示将由线程2打印数字, state=3表示将由线程3打印数字 
    private static int state = 1; 
 
    public static void main(String[] args) { 
        final NumberPrintDemo pn = new NumberPrintDemo(); 
        new Thread(new Runnable() { 
            public void run() { 
                // 3个线程打印75个数字, 单个线程每次打印5个连续数字, 因此每个线程只需执行5次打印任务. 3*5*5=75 
                for (int i = 0; i < 5; i++) { 
                    // 3个线程都使用pn对象做锁, 以保证每个交替期间只有一个线程在打印 
                    synchronized (pn) { 
                        // 如果state!=1, 说明此时尚未轮到线程1打印, 线程1将调用pn的wait()方法, 直到下次被唤醒 
                        while (state != 1) 
                            try { 
                                pn.wait(); 
                            } catch (InterruptedException e) { 
                                e.printStackTrace(); 
                            } 
                        // 当state=1时, 轮到线程1打印5次数字 
                        for (int j = 0; j < 5; j++) { 
                            // 打印一次后n自增 
                            System.out.println(Thread.currentThread().getName() 
                                    + ": " + n++); 
                        } 
                        System.out.println(); 
                        // 线程1打印完成后, 将state赋值为2, 表示接下来将轮到线程2打印 
                        state = 2; 
                        // notifyAll()方法唤醒在pn上wait的线程2和线程3, 同时线程1将退出同步代码块, 释放pn锁.  
                        // 因此3个线程将再次竞争pn锁 
                        // 假如线程1或线程3竞争到资源, 由于state不为1或3, 线程1或线程3将很快再次wait, 释放出刚到手的pn锁.  
                        // 只有线程2可以通过state判定, 所以线程2一定是执行下次打印任务的线程. 
                        // 对于线程2来说, 获得锁的道路也许是曲折的, 但前途一定是光明的. 
                        pn.notifyAll(); 
                    } 
                } 
            } 
        }, "线程1").start(); 
 
        new Thread(new Runnable() { 
            public void run() { 
                for (int i = 0; i < 5; i++) { 
                    synchronized (pn) { 
                        while (state != 2) 
                            try { 
                                pn.wait(); 
                            } catch (InterruptedException e) { 
                                e.printStackTrace(); 
                            } 
                        for (int j = 0; j < 5; j++) { 
                            System.out.println(Thread.currentThread().getName() 
                                    + ": " + n++); 
                        } 
                        System.out.println(); 
                        state = 3; 
                        pn.notifyAll(); 
                    } 
                } 
            } 
        }, "线程2").start(); 
 
        new Thread(new Runnable() { 
            public void run() { 
                for (int i = 0; i < 5; i++) { 
                    synchronized (pn) { 
                        while (state != 3) 
                            try { 
                                pn.wait(); 
                            } catch (InterruptedException e) { 
                                e.printStackTrace(); 
                            } 
                        for (int j = 0; j < 5; j++) { 
                            System.out.println(Thread.currentThread().getName() 
                                    + ": " + n++); 
                        } 
                        System.out.println(); 
                        state = 1; 
                        pn.notifyAll(); 
                    } 
                } 
            } 
        }, "线程3").start(); 
    } 
} 

解法二: 采用JDK1.5并发包提供的Lock, Condition等类的相关方法控制线程.

Java代码 
public class NumberPrint implements Runnable { 
    private int state = 1; 
    private int n = 1; 
    // 使用lock做锁 
    private ReentrantLock lock = new ReentrantLock(); 
    // 获得lock锁的3个分支条件 
    private Condition c1 = lock.newCondition(); 
    private Condition c2 = lock.newCondition(); 
    private Condition c3 = lock.newCondition(); 
 
    @Override 
    public void run() { 
        new Thread(new Runnable() { 
            public void run() { 
                for (int i = 0; i < 5; i++) { 
                    try { 
                        // 线程1获得lock锁后, 其他线程将无法进入需要lock锁的代码块. 
                        // 在lock.lock()和lock.unlock()之间的代码相当于使用了synchronized(lock){} 
                        lock.lock(); 
                        while (state != 1) 
                            try { 
                                // 线程1竞争到了lock, 但是发现state不为1, 说明此时还未轮到线程1打印.  
                                // 因此线程1将在c1上wait 
                                // 与解法一不同的是, 三个线程并非在同一个对象上wait, 也不由同一个对象唤醒 
                                c1.await(); 
                            } catch (InterruptedException e) { 
                                e.printStackTrace(); 
                            } 
                        // 如果线程1竞争到了lock, 也通过了state判定, 将执行打印任务 
                        for (int j = 0; j < 5; j++) { 
                            System.out.println(Thread.currentThread().getName() 
                                    + ": " + n++); 
                        } 
                        System.out.println(); 
                        // 打印完成后将state赋值为2, 表示下一次的打印任务将由线程2执行 
                        state = 2; 
                        // 唤醒在c2分支上wait的线程2 
                        c2.signal(); 
                    } finally { 
                        // 打印任务执行完成后需要确保锁被释放, 因此将释放锁的代码放在finally中 
                        lock.unlock(); 
                    } 
                } 
            } 
        }, "线程1").start(); 
 
        new Thread(new Runnable() { 
            public void run() { 
                for (int i = 0; i < 5; i++) { 
                    try { 
                        lock.lock(); 
                        while (state != 2) 
                            try { 
                                c2.await(); 
                            } catch (InterruptedException e) { 
                                e.printStackTrace(); 
                            } 
                        for (int j = 0; j < 5; j++) { 
                            System.out.println(Thread.currentThread().getName() 
                                    + ": " + n++); 
                        } 
                        System.out.println(); 
                        state = 3; 
                        c3.signal(); 
                    } finally { 
                        lock.unlock(); 
                    } 
                } 
            } 
        }, "线程2").start(); 
 
        new Thread(new Runnable() { 
            public void run() { 
                for (int i = 0; i < 5; i++) { 
                    try { 
 
                        lock.lock(); 
                        while (state != 3) 
                            try { 
                                c3.await(); 
                            } catch (InterruptedException e) { 
                                e.printStackTrace(); 
                            } 
                        for (int j = 0; j < 5; j++) { 
                            System.out.println(Thread.currentThread().getName() 
                                    + ": " + n++); 
                        } 
                        System.out.println(); 
                        state = 1; 
                        c1.signal(); 
                    } finally { 
                        lock.unlock(); 
                    } 
                } 
            } 
        }, "线程3").start(); 
    } 
     
    public static void main(String[] args) { 
        new NumberPrint().run(); 
    } 
} 

总结: 对比解法一和解法二, 显然解法二是更好的解决方案. 解法一的问题在于无法进行精确唤醒, 比如线程1执行完打印任务并调用pn.notifyAll()方法后, 3个线程将再次竞争锁, 而不是精确唤醒线程2. 虽然线程2最终将赢得锁, 下一次的打印任务也肯定会由线程2执行, 但是竞争的持续时间是不可预知的, 只能看线程2的人品.
最糟糕的情形可以是: 线程3竞争到了锁, 紧接着wait. 接下来线程1也竞争到了锁, 然后线程1也wait. 此时就再也没有其他线程跟线程2竞争了, 线程2终于艰难的赢得了锁...

留下3个问题供有兴趣的朋友思考:
1. 解法一和解法二中的while (state != xx)是否可以换成if(state != xx), 为什么?
2. 解法一的中的pn.notifyAll()是否可以换成pn.notify(), 为什么?
3. 是否可以用wait(), notify(), notifyAll()等方法完成类似解法二的精确唤醒, 请给出方案或代码.--这个问题我思考了很久, 却没有头绪. 关键的困难在于必须调用pn的wait()方法和notifyAll()方法, 而不能是其他对象的wait()和notifyAll()方法. 如果你有思路, 还望在博客中留言, 不甚感激!