如何使用同步方法？

Question

我有这段代码.

public class Main{

    public static void main(String[] args) {
    Main1 main1 = new Main1();
    Main2 main2 = new Main2();
    Thread t1 = new Thread(main1);
    Thread t2 = new Thread(main2);
    t1.start();
    t2.start();
  }
}
public class Main1 extends Thread{
    SynchronizedCounter c = new SynchronizedCounter();
    public void run(){
        System.out.println("Entering Main1");
        c.print();
        System.out.println("Exiting Main1");
    }
}
public class Main2 extends Thread{
    SynchronizedCounter c = new SynchronizedCounter();
    public void run(){
        System.out.println("Entering Main2");
        c.print();
        System.out.println("Exiting Main2");
    }
}
public class SynchronizedCounter {
    public void print() {
        for(int  i = 1; i<10; i++){
            System.out.print(i+ " ");
        }
    }
}

当我运行此代码时,我得到不同的输出,我理解为什么.为了使它同步,我在SynchronizedCounter类中将synchronized关键字添加到print().现在,它看起来像这样.

public class SynchronizedCounter {
public synchronized void print() {
    for(int  i = 1; i<10; i++){
        System.out.print(i+ " ");
    }
  }
}

但是,当我运行它时,我得到随机结果.我不确定如何通过使方法同步来使其同步,因此,它给出了固定的结果.

例如 :

Entering Main1
1 2 3 4 5 6 7 8 9 
Exiting Main1
Entering Main2
1 2 3 4 5 6 7 8 9 
Exiting Main2

编辑:谢谢Kayaman的答案.现在工作正常.我只是在代码中稍微改变一下.现在看起来像这样.

public class Main{

public static void main(String[] args) {
    SynchronizedCounter c = new SynchronizedCounter();
    Main1 main1 = new Main1(c);
    Main2 main2 = new Main2(c);
    Thread t1 = new Thread(main1);
    Thread t2 = new Thread(main2);
    t1.start();
    t2.start();
 }
}

public class Main1 extends Thread{
SynchronizedCounter c;
Main1(SynchronizedCounter counter){
    c = counter;
}
public void run(){
    c.print();
 }
}

public class Main2 extends Thread{
SynchronizedCounter c;
Main2(SynchronizedCounter counter){
    c = counter;
}
public void run(){
    c.print();
 }
}

public class SynchronizedCounter {
public static int i = 0;
public synchronized void print() {
    System.out.println("\n Thread "+ i);
    i++;
    for(int  i = 1; i<10; i++){
        System.out.print(i+ " ");
    }
 }
}

现在,输出结果是:

 Thread 0
1 2 3 4 5 6 7 8 9 
 Thread 1
1 2 3 4 5 6 7 8 9 

Answer 1

线程需要使用相同的SynchronizedCounter实例进行同步才能工作.由于代码中有2个实例,因此两个线程都可以在自己的实例上进行同步而不会阻塞.