java threads

[Asanka Vithanage]

the following code gives 1 as output
please is there any body can explin the logic and correct thinking pattern to get that answer

class B extends Thread {
B(){}
B(B a){}
public void run(){
System.out.println("1");
}
}


class A extends B {
A(){}
A(B a){}
public void run(){
System.out.println("2");

}

public static void main( String as[]){

Thread a =new B (new A());

a.start();


}
}

[Rakesh Rajagopalan]

Hi,

B extends Thread and A extends B. Hence, A IS-A B and B IS-A Thread. When you say: Thread a = new B(new A());, JVM chooses Object B at runtime and constructs a new B. Since B's constructor can take a 'B' Object and A IS-A B, therefore the run() method of B is called and the output is 1.

[Jesper de Jong]

Welcome to JavaRanch. Please use code tags when you post code.

[Ankit Garg]

Asanka this is because the internal implementation of the run method in Thread class. The internal implementation of Thread class looks like this

class Thread implements Runnable
{
   Runnable runner;
   Thread(Runnable runnable)
   {
      runner = runnable;
   }
   public void run()
   {
      if(runner != null)
      {
         runner.run();
      }
      //some other code (I think it calls notifyAll here but this is not relevant here)
   }
}

So when you use this code

class MyRunnable implements Runnable
{
   public void run()
   {
      System.out.println("2");
   }
   public static void main(String[] args)
   {
      Thread t = new Thread(new MyRunnable());
      t.start();
   }
}

The run method of Thread class is called and it calls the run method of MyRunnable class.

But in you code the run method of B will be called and it doesn't calls the run method of A class. Infact, it wastes the Runnable passed to it as a parameter

B(B a)
{//does nothing}

This constructor declaration should have been like this

B(Runnable r)
{
super(r);
}