如何使用多线程打印序列中的数字

Question

我想用 T1-1、T2-2、T3-3、T1-4、T2-5、T3-6 等线程按顺序打印数字public class NumberGame {    static int a=1;    public static void main(String args[]) throws InterruptedException    {        PrintSequenceRunnable C1=new PrintSequenceRunnable("T1",a);        PrintSequenceRunnable C2=new PrintSequenceRunnable("T2",a);        PrintSequenceRunnable C3=new PrintSequenceRunnable("T3",a);        Thread t1 = new Thread(C1);        Thread t2 = new Thread(C2);        Thread t3 = new Thread(C3);        t1.start();        t2.start();        t3.start();     }}public class PrintSequenceRunnable implements Runnable {    String tname;    int a;    PrintSequenceRunnable(String tname, int a )    {        this.tname = tname;        this.a = a;    }    @Override    public void run() {        synchronized (this) {            for(int i=0; i<10;i++)            {                System.out.println(tname+" "+a);                a++;                try {                    this.wait(1000);                    this.notify();                } catch (InterruptedException e) {                    // TODO Auto-generated catch block                    e.printStackTrace();                }            }        }        // TODO Auto-generated method stub    }}但我的输出就像T1-1 T2-1 T3-1 T1-2 T3-2 T2-2 T3-3 T1-3 T2-3 T3-4 T1-4 T2-4 T3-5 T1-5 T2-5 T3-6 T1- 6 T2-6 T1-7 T2-7 T3-7 T2-8 T3-8 T1-8 T2-9 T3-9 T1-9 T2-10 T1-10 T3-10谁能帮我。

Answer 1

问题是：

1. 该设计存在线程之间引发条件的问题，您需要同步它们。

2. 在使用构造函数时，PrintSequenceRunnable(String tname, int a )您将发送原始变量的副本a，它是静态成员NumberGame。所以，每个PrintSequenceRunnable都有自己的变量a。

我的建议是使用方法wait和同步每个线程notify。我拿了你的代码并做了一些修改：

数字游戏

public class NumberGame {

    public static void main(String args[]) throws InterruptedException

    {       

        PrintSequenceRunnable C1=new PrintSequenceRunnable("T1");

        PrintSequenceRunnable C2=new PrintSequenceRunnable("T2");

        PrintSequenceRunnable C3=new PrintSequenceRunnable("T3");

        Thread t1 = new Thread(C1);

        Thread t2 = new Thread(C2);

        Thread t3 = new Thread(C3);

        t1.start();

        t2.start();

        t3.start();

        Thread.sleep(1);//Wait 1 ms to avoid a raise condition

        PrintSequenceRunnable.activateNextItem(); //Start sequence.

        t1.join();

        t2.join();

        t3.join();

        System.out.println("--END--");

    }

}

打印序列可运行

import java.util.Vector;

public class PrintSequenceRunnable implements Runnable {    

    static private int a = 0;

    private static Vector<PrintSequenceRunnable> items = new Vector<PrintSequenceRunnable>();

    /**

     * Method to select the next Thread which will be activate to continue its thread.

     */

    public static synchronized void activateNextItem() {

        int index = a % items.size();

        items.get(index).activate();

    }

    private String tname;

    private Object sempahoro = new Object(); //Object to sinchrony the  thread

    public PrintSequenceRunnable(String tname)

    {

        this.tname = tname;

        items.add(this);

    }

    public void activate()

    {

        synchronized (sempahoro) {

            sempahoro.notify();

        }

    }

    @Override

    public void run() {

        for(int i=0; i<10;i++)

        {

            synchronized (sempahoro) {

                try {

                    sempahoro.wait();

                } catch (InterruptedException e) {

                    e.printStackTrace();

                }

            }

            a++;

            System.out.println(tname+" "+a);            

            activateNextItem(); //Raise the next thread.

        }

        // TODO Auto-generated method stub

    }

}

在此示例中，方法activateNextItem, fromPrintSequenceRunnable` 将决定通知哪个实例执行其线程。

重要的是，我需要sleep在初始化每个线程后设置一秒以避免引发条件，我的意思是：等待所有线程启动并使所有线程处于等待状态。

输出：

T1 1

T2 2

T3 3

T1 4

T2 5

T3 6

T1 7

T2 8

T3 9

T1 10

T2 11

T3 12

T1 13

T2 14

T3 15

T1 16

T2 17

T3 18

T1 19

T2 20

T3 21

T1 22

T2 23

T3 24

T1 25

T2 26

T3 27

T1 28

T2 29

T3 30

--END--

Answer 2

只是为了学习，这就是强制顺序输出的方法。但请注意，不能以任何方式保证线程的顺序执行。正如其他人指出的那样，这个问题不适合多线程处理。如果您想按顺序执行某件事，请在一个线程中执行。

public class NumberGame {

    public static void main(String[] args) {

        PrintSequenceRunnable.startFrom("T1");

        new Thread(new PrintSequenceRunnable("T1", "T2")).start();

        new Thread(new PrintSequenceRunnable("T2", "T3")).start();

        new Thread(new PrintSequenceRunnable("T3", "T1")).start();

    }

}

class PrintSequenceRunnable implements Runnable {

    private final String name;

    private final String next;

    private static String moveTo;

    private static int value = 1;

    PrintSequenceRunnable(String name, String next) {

        this.name = name;

        this.next = next;

    }

    static void startFrom(String start) {

        moveTo = start;

    }

    private int uselessCounter = 0;

    @Override

    public void run() {

        do {

            synchronized (moveTo) {

                if (name.equals(moveTo)) {

                    System.out.println(name + "-" + (value++));

                    moveTo = next;

                } else {

                    uselessCounter++;

                }

            }

        } while (value < 10);

        System.out.println("Ran " + name + " uselessly for " + uselessCounter + " times."); // remove it.

    }

}