Question

java线程问题：怎样多次执行

Answer 1

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();  
}  
}