学习Java中的线程,在理解某些方面时遇到困难

Question

我正在网上学习Java的线程,我是初学者,我在理解一些概念方面遇到了困难.谁能帮我？

链接我正在学习:http://www.codeproject.com/Articles/616109/Java-Thread-Tutorial

问题:

public class Core {
    public static void main(String[] args) {
        Runnable r0,r1;//pointers to a thread method
        r0=new FirstIteration("Danial"); //init Runnable, and pass arg to thread 1
        SecondIteration si=new SecondIteration();
        si.setArg("Pedram");// pass arg to thread 2

        r1=si; //<~~~ What is Happening here??

        Thread t0,t1;
        t0=new Thread(r0);
        t1=new Thread(r1);
        t0.start();
        t1.start();
        System.out.print("Threads started with args, nothing more!\n");
    }
}

编辑:FirstIteration和SceondIteration的代码

class FirstIteration implements Runnable{
    public FirstIteration(String arg){
        //input arg for this class, but in fact input arg for this thread.
        this.arg=arg;
    }

    private String arg;

    @Override
    public void run() {
        for(int i=0;i<20;i++){
        System.out.print("Hello from 1st. thread, my name is "+
            arg+"\n");//using the passed(arg) value
        }
    }
}

class SecondIteration implements Runnable{
    public void setArg(String arg){//pass arg either by constructors or methods.
        this.arg=arg;
    }

    String arg;

    @Override
    public void run() {
        for(int i=0;i<20;i++){
            System.out.print("2)my arg is="+arg+", "+i+"\n");
        }
    }
}

Answer 1

R1 = SI; <~~~这里发生了什么？*

参考作业.

您正在创建的参考副本si在r1.因此,该声明后,双方r1并si会指向同一个对象.

            ______________________
si ------> |new SecondIteration();|   // You have 1 reference to this object
           ------------------------ 

r1 = si;  // Make Runnable reference `r1` point the same object that `si` refers

            ______________________
si ------> |new SecondIteration();|   // Now you have 2 reference to this object
           ------------------------
            ^
            |
r1 ----------

实际上,他们可以简单地完成:

r1 = new SecondIteration();

而不是那两个步骤.但是,因为setArg()方法未在Runnable接口中声明.所以,要调用它,你必须进行类型转换,如下所示:

((SecondIteration)r1).setArg();

这可能不太可读,这就是为什么他们可能将其分为两个步骤:

SecondIteration si = new SecondIteration();
si.setArg("Pedram"); 
r1=si;