public class Quest implements Runnable
{
    int b = 100;

    public synchronized void m1() throws Exception
    {
        System.out.println("enter m1");
        b = 1000;
        Thread.sleep(3000);
        System.out.println("b="+b);
    }
    
    public void m2() throws Exception
    {
        System.out.println(
 
"enter m2");
        Thread.sleep(3000);
        b = 2000;
    }
    

    @Override
    public void run()
    {
            try
            {
                m1();
            } catch (Exception e)
            {
                e.printStackTrace();
            }
    }
    
    public static
 
 void main(String[] args)
    {
        try
        {
            Quest quest = new Quest();
            // 執行緒從呼叫start()到run()執行需要一定的時間
            new Thread(quest).start();
            // 所以此處會先執行m2()
            quest.m2();
            System.out.println(quest.b);
        } catch (Exception e)
        {
            e.printStackTrace();
        }
    }

}
 

結果：

　　每次先列印 m2