java - java中原始整数值的同步

Question

我有两个java线程。其中一个打印偶数，另一个打印奇数。我需要按自然顺序打印数字。是否可以将两个线程同步为仅使用原始整数，如下所示？Node:原始赋值在 java 中跨 jvm 是原子的。

public class NaturalNumber{

volatile   int  ai = 0;
public static void main(String str[]){
    final NaturalNumber nn = new NaturalNumber();

        Thread even = new Thread(new Runnable(){
        int i=0;
            public void run(){
                //int i=0;
                while(i<=200){
                    if(nn.ai ==0){
                        System.out.println(i);
                        i=i+2;
                        nn.ai =1 ;

                    }
                }
            }
        }); 

        Thread odd = new Thread( new Runnable(){
            int i=1;
            public void run(){
                //int i=1;
                while(i<=200){
                    if(nn.ai ==1) {
                        System.out.println(i);
                        i=i+2;
                        nn.ai =0 ;
                    }
                }
            }
        });


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


}

}

Answer 1

这是使用锁定和同步来实现它的方法，我认为这不太容易出错。

public class NaturalNumber{

private boolean printEven = true;
private Object lock = new Object();

public NaturalNumber()
{
    new Thread(()->even()).start();
    new Thread(()->odd()).start();
}

private void even()
{
    int i = 0;
    while(i<=200)
    {
        synchronized (lock)
        {
            if(printEven)
            {
                System.out.println(i);
                i += 2;
                lock.notify();
                printEven = false;
            }
            else //flag says next odd is to be printed, so wait until it has
            {
                try
                {
                    lock.wait();
                }
                catch (InterruptedException e)
                {
                    e.printStackTrace();
                }
            }
        }
    }
}
private void odd()
{
    int i = 1;
    while(i<=200)
    {
        synchronized (lock)
        {
            if(!printEven)
            {
                System.out.println(i);
                i += 2;
                lock.notify();
                printEven = true;
            }
            else //flag says next even is to be printed, so wait until it has
            {
                try
                {
                    lock.wait();
                }
                catch (InterruptedException e)
                {
                    e.printStackTrace();
                }
            }
        }
    }
}

public static void main(String str[]){

    new NaturalNumber();
}
}