在 Java 中使用两个线程打印偶数和奇数答案

【问题标题】：Printing Even and Odd using two Threads in Java在 Java 中使用两个线程打印偶数和奇数
【发布时间】：2013-05-17 08:49:58
【问题描述】：

我尝试了下面的代码。我从其他一些帖子中获取了这段代码，根据作者的说法是正确的。但是当我尝试跑步时，它并没有给我确切的结果。

这主要是顺序打印偶数和奇数。

public class PrintEvenOddTester {



    public static void main(String ... args){
        Printer print = new Printer(false);
        Thread t1 = new Thread(new TaskEvenOdd(print));
        Thread t2 = new Thread(new TaskEvenOdd(print));
        t1.start();
        t2.start();
    }


}



class TaskEvenOdd implements Runnable {

    int number=1;
    Printer print;

    TaskEvenOdd(Printer print){
        this.print = print;
    }

    @Override
    public void run() {

        System.out.println("Run method");
        while(number<10){

            if(number%2 == 0){
                System.out.println("Number is :"+ number);
                print.printEven(number);
                number+=2;
            }
            else {
                System.out.println("Number is :"+ number);
                print.printOdd(number);
                number+=2;
            }
        }

      }

    }

class Printer {

    boolean isOdd;

    Printer(boolean isOdd){
        this.isOdd = isOdd;
    }

    synchronized void printEven(int number) {

        while(isOdd){
            try {
                wait();
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }
        System.out.println("Even:"+number);
        isOdd = true;
        notifyAll();
    }

    synchronized void printOdd(int number) {
        while(!isOdd){
            try {
                wait();
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }
        System.out.println("Odd:"+number);
        isOdd = false;
        notifyAll();
    }

}

有人可以帮我解决这个问题吗？

编辑预期结果：奇数：1 偶数：2 奇数：3 偶数：4 奇数：5 偶数：6 奇数：7 偶数：8 奇数：9

【问题讨论】：

实际结果是什么，您的预期结果是什么？
数字从 1 开始，你只能将它增加 2。因此它永远不会是偶数。
这不是调试服务...
对于每一位来到这里的学生：请告诉你的导师，虽然这个练习可能会教你一些关于如何控制线程的知识，但这是一个非常可怕的示例为什么 使用线程。如果您希望程序以特定顺序（例如，1、2、3，...）执行某些事情（例如，打印数字）；那么绝对最好的方法就是在一个线程中完成这些事情。每个多线程程序都需要线程之间的一些同步，但是你使用的同步越多，你从线程中获得的好处就越少。这个程序实际上得到了负面的好处。
没有理由让线程交替打印数字，甚至撇开关于增量的错误。您的期望是误入歧途的，未引用的“其他一些代码”也是如此。 SO 不是针对任意 Internet 垃圾的验证服务。

标签： java multithreading synchronization

【解决方案1】：

此解决方案适用于我（Java 8 及更高版本使用 Lamda 表达式）

public class EvenOddUsingThread {

    static int number = 100;
    static int counter = 1;

    public static void main(String[] args) {
        EvenOddUsingThread eod = new EvenOddUsingThread();
        Thread t1 = new Thread(() -> eod.printOdd());
        Thread t2 = new Thread(() -> eod.printEven());
        t1.start();
        t2.start();
    }

    public void printOdd() {
        synchronized (this) {
            while (counter < number) {
                if (counter % 2 == 0) {
                    try {
                        wait();
                    } catch (InterruptedException e) {
                    }
                }
                // Print the number
                System.out.println(counter + " Thread Name: " + Thread.currentThread().getName());
                // Increment counter
                counter++;
                // Notify to second thread
                notify();
            }
        }
    }

    public void printEven() {
        synchronized (this) {
            while (counter < number) {
                if (counter % 2 == 1) {
                    try {
                        wait();
                    } catch (InterruptedException ignored) {
                    }
                }
                // Print the number
                System.out.println(counter + " Thread Name: " + Thread.currentThread().getName());
                // Increment counter
                counter++;
                // Notify to second thread
                notify();
            }
        }
    }

}

【讨论】：

【解决方案2】：

下面的解决方案是使用 java 8 可完成的 future 和 executor 服务使用两个线程打印偶数和奇数。

ExecutorService firstExecutorService = Executors.newSingleThreadExecutor(r -> {
        Thread t = new Thread(r);
        t.setName("first");
        return t;
    });

    ExecutorService secondExecutorService = Executors.newSingleThreadExecutor(r -> {
        Thread t = new Thread(r);
        t.setName("second");
        return t;
    });

    IntStream.range(1, 101).forEach(num -> {

        CompletableFuture<Integer> thenApplyAsync = CompletableFuture.completedFuture(num).thenApplyAsync(x -> {

            if (x % 2 == 1) {
                System.out.println(x + " " + Thread.currentThread().getName());
            }
            return num;
        }, firstExecutorService);

        thenApplyAsync.join();

        CompletableFuture<Integer> thenApplyAsync2 = CompletableFuture.completedFuture(num).thenApplyAsync(x -> {
            if (x % 2 == 0) {
                System.out.println(x + " " + Thread.currentThread().getName());
            }
            return num;
        }, secondExecutorService);

        thenApplyAsync2.join();
    });

    firstExecutorService.shutdown();
    secondExecutorService.shutdown();

下面是它的控制台日志。

【讨论】：

【解决方案3】：

公共类 EvenOddex {

public static class print {

    int n;
    boolean isOdd = false;

    synchronized public void printEven(int n) {

        while (isOdd) {
            try {
                wait();
            } catch (InterruptedException ex) {
                Logger.getLogger(EvenOddex.class.getName()).log(Level.SEVERE, null, ex);
            }

        }

        System.out.print(Thread.currentThread().getName() + n + "\n");

        isOdd = true;
        notify();
    }

    synchronized public void printOdd(int n) {
        while (!isOdd) {
            try {
                wait();
            } catch (InterruptedException ex) {
                Logger.getLogger(EvenOddex.class.getName()).log(Level.SEVERE, null, ex);
            }
        }


        System.out.print(Thread.currentThread().getName() + n + "\n");
        isOdd = false;
        notify();



    }
}

public static class even extends Thread {

    print po;

    even(print po) {

        this.po = po;

        new Thread(this, "Even").start();

    }

    @Override
    public void run() {


        for (int j = 0; j < 10; j++) {
            if ((j % 2) == 0) {
                po.printEven(j);
            }
        }

    }
}

public static class odd extends Thread {

    print po;

    odd(print po) {

        this.po = po;
        new Thread(this, "Odd").start();
    }

    @Override
    public void run() {

        for (int i = 0; i < 10; i++) {

            if ((i % 2) != 0) {
                po.printOdd(i);
            }
        }

    }
}

public static void main(String args[]) {
    print po = new print();
    new even(po);
    new odd(po);

}

}

【讨论】：

【解决方案4】：

public class Multi extends Thread{  
    public static int a;
    static{a=1;}
    public void run(){  
        for(int i=1;i<5;i++){  
        System.out.println("Thread Id  "+this.getId()+"  Value "+a++);
        try{Thread.sleep(500);}catch(InterruptedException e){System.out.println(e);}  

        }  
    }  
public static void main(String args[]){  
       Multi t1=new Multi();  
       Multi t2=new Multi();  

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

【讨论】：

程序的正确性取决于'Thread.sleep();'你在那里写的。它会产生不可预知的结果。我们的想法是创建一个可以确定地产生预期结果的程序。

【解决方案5】：

public class PrintOddEven {
private static class PrinterThread extends Thread {

    private static int current = 0;
    private static final Object LOCK = new Object();

    private PrinterThread(String name, int number) {
        this.name = name;
        this.number = number;
    }

    @Override
    public void run() {
        while (true) {
            synchronized (LOCK) {
                try {
                    LOCK.wait(1000);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }

                if (current < number) {
                    System.out.println(name + ++current);
                } else {
                    break;
                }

                LOCK.notifyAll();
            }
        }
    }

    int number;
    String name;
}

public static void main(String[] args) {
    new PrinterThread("thread1 : ", 20).start();
    new PrinterThread("thread2 : ", 20).start();
}
}

【讨论】：

【解决方案6】：

公共类解决方案{

 static class NumberGenerator{

     private static volatile boolean printEvenNumber = false;


     public  void printEvenNumber(int i) {
         synchronized (this) {
             if(!printEvenNumber) {
                 try {
                     wait();
                 } catch (InterruptedException e) {
                     e.printStackTrace();
                 }
             }
             System.out.println(i);
             printEvenNumber = !printEvenNumber;
             notify();
         }
     }

     public  void printOddNumber(int i ) {
            synchronized (this) {
                if(printEvenNumber) {
                    try {
                        wait();
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                }

                System.out.println(i);
                printEvenNumber = !printEvenNumber;
                notify();
            }
     }

}

static  class OddNumberGenerator implements Runnable{
    private NumberGenerator numberGenerator;

    public OddNumberGenerator(NumberGenerator numberGenerator) {
        this.numberGenerator = numberGenerator;
    }

    @Override
    public void run() {
        for(int i  = 1; i <100; i = i + 2) {
            numberGenerator.printOddNumber(i);
        }
    }
}

static class EvenNumberGenerator implements Runnable {
    private NumberGenerator numberGenerator;

    public EvenNumberGenerator(NumberGenerator numberGenerator) {
        this.numberGenerator = numberGenerator;
    }

    @Override
    public void run() {
        for (int i = 2; i <= 100; i =  i + 2) {
           numberGenerator.printEvenNumber(i);
        }
    }
}


public static void main(String[] args) {
    NumberGenerator ng = new NumberGenerator();
    OddNumberGenerator oddNumberGenerator = new OddNumberGenerator(ng);
    EvenNumberGenerator evenNumberGenerator = new EvenNumberGenerator(ng);
    new Thread(oddNumberGenerator).start();
    new Thread(evenNumberGenerator).start();

}

}

【讨论】：

【解决方案7】：

package programs.multithreading;

public class PrintOddEvenNoInSequence {

final int upto;
final PrintOddEvenNoInSequence obj;
volatile boolean oddFlag,evenFlag;
public PrintOddEvenNoInSequence(int upto){
    this.upto = upto;
    obj = this;
    oddFlag = true;
    evenFlag = false;
}
void printInSequence(){

    Thread odd = new Thread(new Runnable() {
        @Override
        public void run() {
            for(int i = 1; i <= upto; i = i + 2){
                synchronized (obj) {
                    while(!oddFlag){
                        try {
                            obj.wait();
                        } catch (InterruptedException e) {
                            // TODO Auto-generated catch block
                            e.printStackTrace();
                        }
                    }
                    System.out.println("Odd:"+i);
                    oddFlag = false;
                    evenFlag = true;
                    obj.notify();
                }
            }
        }
    });

    Thread even = new Thread(new Runnable() {
        @Override
        public void run() {
            for(int i = 2; i <= upto; i = i + 2){
                synchronized (obj) {
                    while(!evenFlag){
                        try {
                            obj.wait();
                        } catch (InterruptedException e) {
                            // TODO Auto-generated catch block
                            e.printStackTrace();
                        }
                    }
                    System.out.println("Even:"+i);
                    oddFlag = true;
                    evenFlag = false;
                    obj.notify();
                }
            }
        }
    });

    odd.start();
    even.start();

}
public static void main(String[] args) {
    new PrintOddEvenNoInSequence(100).printInSequence();
}
}

【讨论】：

虽然这段代码 sn-p 可以解决问题，但including an explanation 确实有助于提高帖子的质量。请记住，您是在为将来的读者回答问题，而这些人可能不知道您提出代码建议的原因。

【解决方案8】：

查看 Clean 实现

public class PrintOddEvenByTwoThreads {
    static int number = 1;
    static Thread odd;
    static Thread even;
    static int max = 10;

    static class OddThread extends Thread {
        @Override
        public void run() {
            while (number <= max) {
                if (number % 2 == 1) {
                    System.out.println(Thread.currentThread() + "" + number++);
                } else {

                    synchronized (odd) {
                        synchronized (even) {
                            even.notify();
                        }
                        try {
                            wait();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                }
            }
        }
    }

    static class EvenThread extends Thread {
        @Override
        public void run() {
            while (number <= max) {
                if (number % 2 == 0) {
                    System.out.println(Thread.currentThread() + "" + number++);
                } else {

                    synchronized (even) {
                        synchronized (odd) {
                            odd.notify();
                        }
                        try {
                            wait();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                }
            }
        }
    }

    public static void main(String[] args) throws InterruptedException {
        odd = new OddThread();
        even = new EvenThread();
        odd.start();
        even.start();
    }
}

【讨论】：

【解决方案9】：

1- 数字初始化为 1，isOdd 标志设置为 false。将isOdd 设置为 true

2- 增量应为 1（而不是 2），即 number+=1；

【讨论】：

【解决方案10】：

import java.util.concurrent.Semaphore;


public class PrintOddAndEven {

private static class OddThread extends Thread {
    private Semaphore semaphore;
    private Semaphore otherSemaphore;
    private int value = 1;

    public  OddThread(Semaphore semaphore, Semaphore otherSemaphore) {
        this.semaphore = semaphore;
        this.otherSemaphore = otherSemaphore;
    }

    public void run() {
        while (value <= 100) {
            try {
                // Acquire odd semaphore
                semaphore.acquire();
                System.out.println(" Odd Thread " + value + " " + Thread.currentThread().getName());

            } catch (InterruptedException excetion) {
                excetion.printStackTrace();
            }
            value = value + 2;
            // Release odd semaphore
            otherSemaphore.release();
        }
    }
}


private static class EvenThread extends Thread {
    private Semaphore semaphore;
    private Semaphore otherSemaphore;

    private int value = 2;

    public  EvenThread(Semaphore semaphore, Semaphore otherSemaphore) {
        this.semaphore = semaphore;
        this.otherSemaphore = otherSemaphore;
    }

    public void run() {
        while (value <= 100) {
            try {
                // Acquire even semaphore
                semaphore.acquire();
                System.out.println(" Even Thread " + value + " " + Thread.currentThread().getName());

            } catch (InterruptedException excetion) {
                excetion.printStackTrace();
            }
            value = value + 2;
            // Release odd semaphore
            otherSemaphore.release();
        }
    }
}


public static void main(String[] args) {
    //Initialize oddSemaphore with permit 1
    Semaphore oddSemaphore = new Semaphore(1);
    //Initialize evenSempahore with permit 0
    Semaphore evenSempahore = new Semaphore(0);
    OddThread oddThread = new OddThread(oddSemaphore, evenSempahore);
    EvenThread evenThread = new EvenThread(evenSempahore, oddSemaphore);
    oddThread.start();
    evenThread.start();
    }
}

【讨论】：

在代码中添加了 cmets。程序打印 1 到 100，在两个线程之间共享。

【解决方案11】：

下面的简单解决方案：-

package com.test;

class MyThread implements Runnable{

    @Override
    public void run() {
        int i=1;
        while(true) {
            String name=Thread.currentThread().getName();
            if(name.equals("task1") && i%2!=0) {
                System.out.println(name+"::::"+i);
                try {
                    Thread.sleep(1000);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }else if(name.equals("task2") && i%2==0){
                System.out.println(name+"::::"+i);
                try {
                    Thread.sleep(1000);
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }
            i++;
        }

    }

    public static void main(String[] args) {

        MyThread task1=new MyThread();
        MyThread task2=new MyThread();

        Thread t1=new Thread(task1,"task1");
        Thread t2=new Thread(task2,"task2");

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

    }

}

【讨论】：