What if we call run() method directly instead start() method?

• Each thread starts in a separate call stack.
• Invoking the run() method from main thread, the run() method goes onto the current call stack rather than at the beginning of a new call stack.

class TestCallRun1 extends Thread{  

 public void run(){  

   System.out.println("running...");  

 }  

 public static void main(String args[]){  

  TestCallRun1 t1=new TestCallRun1();  

  t1.run();//fine, but does not start a separate call stack  

 }  

}  

Output:running...

 Problem if you direct call run() method

class TestCallRun2 extends Thread{  

 public void run(){  

  for(int i=1;i<5;i++){  

    try{Thread.sleep(500);}catch(InterruptedException e){System.out.println(e);}  

    System.out.println(i);  

  }  

 }  

 public static void main(String args[]){  

  TestCallRun2 t1=new TestCallRun2();  

  TestCallRun2 t2=new TestCallRun2();  

   

  t1.run();  

  t2.run();  

 }  

}  

Output:1
       2

       4
3
       5

       3
1
       2
       4

       5




As you can see in the above program that there is no context-switching because here t1 and t2 will be treated as normal object not thread object.