具有三个线程的Java线程协作打印额外的数字

Question

我试图通过三个线程打印三个AP序列(增量为3)，如下所示:

• 线程-1正在打印序列1、4、7、10、
• 线程2正在打印序列2、5、8、…
• 线程-3正在打印序列3、6、9、…

当任何其他线程轮流打印其序列的数字时，线程应该等待。线程应该相互合作，从1到LIMIT（一个整数；这里LIMIT=10）依次打印数字。

（对于极限 = 10）

1 (printed by Thread-1)
2 (printed by Thread-2)
3 (printed by Thread-3)
4 (printed by Thread-1)
5 (printed by Thread-2)
6 (printed by Thread-3)
7 (printed by Thread-1)
8 (printed by Thread-2)
9 (printed by Thread-3)
10 (printed by Thread-1)

（对于极限 = 10）

1 (printed by Thread-1)
2 (printed by Thread-2)
3 (printed by Thread-3)
4 (printed by Thread-1)
5 (printed by Thread-2)
6 (printed by Thread-3)
7 (printed by Thread-1)
8 (printed by Thread-2)
9 (printed by Thread-3)
10 (printed by Thread-1)
11 (printed by Thread-2)
12 (printed by Thread-3)

class PrintingSequences {
    private static final int LIMIT = 10;
    int counter = 1;
    boolean isPrinting = false;

    // prints 1, 4, 7, 10, ...
    synchronized void printAPStartingFrom1() {
        while (counter <= LIMIT) {
            while (isPrinting || counter % 3 != 1) {
                try {
                    wait();
                } catch (InterruptedException e) {
                    System.out.println(e);
                    Thread.currentThread().interrupt();
                }
            }
            isPrinting = true;
            System.out.println(counter++ + " (printed by " + Thread.currentThread().getName() + ")");
            isPrinting = false;
            notifyAll();
        }
    }

    // prints 2, 5, 8, 11, ...
    synchronized void printAPStartingFrom2() {
        while (counter <= LIMIT) {
            while (isPrinting || counter % 3 != 2) {
                try {
                    wait();
                } catch (InterruptedException e) {
                    System.out.println(e);
                    Thread.currentThread().interrupt();
                }
            }
            isPrinting = true;
            System.out.println(counter++ + " (printed by " + Thread.currentThread().getName() + ")");
            isPrinting = false;
            notifyAll();
        }
    }

    // prints 3, 6, 9, 12, ...
    synchronized void printAPStartingFrom3() {
        while (counter <= LIMIT) {
            while (isPrinting || counter % 3 != 0) {
                try {
                    wait();
                } catch (InterruptedException e) {
                    System.out.println(e);
                    Thread.currentThread().interrupt();
                }
            }
            isPrinting = true;
            System.out.println(counter++ + " (printed by " + Thread.currentThread().getName() + ")");
            isPrinting = false;
            notifyAll();
        }
    }
}

public class TripleThreadCommunication {
    public static void main(String[] args) {
        PrintingSequences naturalNumbers = new PrintingSequences();

        new Thread("Thread-1") {
            @Override
            public void run() {
                naturalNumbers.printAPStartingFrom1();
            }
        }.start();

        new Thread("Thread-2") {
            @Override
            public void run() {
                naturalNumbers.printAPStartingFrom2();
            }
        }.start();

        new Thread("Thread-3") {
            @Override
            public void run() {
                naturalNumbers.printAPStartingFrom3();
            }
        }.start();
    }
}

该程序有三种不同的同步方法：printAPStartingFrom1（）、printAPStartingFrom2（）、printAPStartingFrom3（），分别由Thread-1、Thread-2和Thread-3调用。线程使用waiting（）和elle fyAll（）方法相互配合。

为甚么结果总是涉及两个额外数字，即11和12，而这两个数字又超出了既定的10个上限？

Answer 1

当线程被唤醒时(< code>wait)

我建议您提前确定每个线程将循环多少次(这将使代码更简单):

    // thread1
    // prints 1, 4, 7, 10, ...
    synchronized void printAPStartingFrom1() {
        int count = LIMIT % 3 == 0 ? LIMIT / 3 : LIMIT / 3 + 1;
        for (int i = 0; i < count; i++) {
            while (counter % 3 != 1) {
                wait();
            }
            printAndAddCounter();
            notifyAll();
        }
    }
    // thread2
    // prints 2, 5, 8, 11, ...
    synchronized void printAPStartingFrom1() {
        int count = (LIMIT - 1) % 3 == 0 ? LIMIT / 3 : LIMIT / 3 + 1;
        for (int i = 0; i < count; i++) {
            while (counter % 3 != 2) {
                wait();
            }
            printAndAddCounter();
            notifyAll();
        }
    }
    // thread3
    // prints 3, 6, 9, 12, ...
    synchronized void printAPStartingFrom1() {
        int count = LIMIT / 3;
        for (int i = 0; i < count; i++) {
            while (counter % 3 != 0) {
                wait();
            }
            printAndAddCounter();
            notifyAll();
        }
    }