Question

我已经开始在java中学习线程了。我已经编写了一个代码来按顺序打印奇数和偶数。但我得到的输出是两次。以下是我的代码。

ThreadBasic1.java：

public class ThreadBasic1 {

    public static void main(String[] args) {


       Thread t1 = new Thread(new ThreadImplementation1());
       Thread t2 = new Thread(new ThreadImplementation1());

       t1.start();
       t2.start();

    }
}

ThreadImplementation1.java：

public class ThreadImplementation1 implements Runnable {

    boolean isOdd= false;
    @Override
    public void run() {
        System.out.println("thread started :: Thread name :: " + Thread.currentThread());
        try {
           for(int i = 1 ; i < 10 ; i++) {
               if(i %2 ==0) {
                   printEven(i);
               }
               else {
                   printOdd(i);
               }
           }
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
    }





    synchronized void printEven(int number)throws InterruptedException  {
        while(!isOdd){
            try {

                wait();

            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }
        System.out.println("Even:"+number);
        isOdd = false;
        notifyAll();
    }
    synchronized void printOdd(int number) throws InterruptedException {

        while (isOdd) {

            try {
                wait();
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }

        System.out.println("Odd :"+number);

        isOdd = true;
        notifyAll();
    }

}

输出：

thread started :: Thread name :: Thread[Thread-1,5,main]
Odd :1
Even:2
Odd :3
Even:4
Odd :5
Even:6
Odd :7
Even:8
Odd :9
thread started :: Thread name :: Thread[Thread-0,5,main]
Odd :1
Even:2
Odd :3
Even:4
Odd :5
Even:6
Odd :7
Even:8
Odd :9

Answer 1

你的问题很简单：你有两个完全相同的线程：打印从1到9的数字。

通常，此类练习要求您为每个线程设置略有不同的代码。一个线程仅打印奇数，其他线程打印偶数。然后两个线程都需要进入锁定步骤，以实现正确的1 2 3 4输出。或者你有“一个”线程实现，但你以某种方式启用配置该线程的行为。

因此，这里的答案是退后一步，重新思考整个方法。让两个线程执行相同是没有意义的。这只是意味着你重复某些行为。相反，想象一下如何对齐两个不同的行为。

当然，我不会给你代码：练习的重点是你将这些想法变成代码。

Answer 2

您可以在这里使用余数的概念。

如果number％2 == 1，则Odd将打印该数字并递增，否则它将进入等待状态。如果number％2 == 0，则偶数将打印该数字并递增，否则它将进入等待状态。让我们借助示例进行检查。

创建一个名为“ OddEvenRunnable”的类并实现Runnable接口。

class OddEvenRunnable implements Runnable {
    public int PRINT_NUMBERS_UPTO = 10;
    static int number = 1;
    int remainder;
    static Object lock = new Object();


    public OddEvenRunnable(int remainder) {
        this.remainder = remainder;
    }

    @Override
    public void run() {
        while (number < PRINT_NUMBERS_UPTO) {
            synchronized (lock) {
                while (number % 2 != remainder) {
                    try {
                        lock.wait();
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                }
                System.out.println(Thread.currentThread().getName() + " " + number);
                number++;
                lock.notifyAll();
            }
        }

    }
}

public class PrintOddEvenMain {

    public static void main(String[] args) {
        OddEvenRunnable oddRunnable = new OddEvenRunnable(1);
        OddEvenRunnable evenRunnable = new OddEvenRunnable(0);


        Thread t1 = new Thread(oddRunnable, "Odd");
        Thread t2 = new Thread(evenRunnable, "Even");

        t1.start();
        t2.start();
    }

}

Answer 3

理想情况下，你应该只运行一个线程来执行切换作业b / w偶数和奇数。

package com.samples;

public class ThreadBasic1 {
public static void main(String[] args) {
    Thread t1 = new Thread(new ThreadImplementation1());
    Thread t2 = new Thread(new ThreadImplementation1());

    t1.start();
    //t2.start();
  }
}

Java多线程打印奇数和偶数

3 个答案: