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.

1. class TestCallRun1 extends Thread{  
2.  public void run(){  
3.    System.out.println("running...");  
4.  }  
5.  public static void main(String args[]){  
6.   TestCallRun1 t1=new TestCallRun1();  
7.   t1.run();//fine, but does not start a separate call stack  
8.  }  
9. }  

Output:running...

Problem if you direct call run() method

1. class TestCallRun2 extends Thread{  
2.  public void run(){  
3.   for(int i=1;i<5;i++){  
4.     try{Thread.sleep(500);}catch(InterruptedException e){System.out.println(e);}  
5.     System.out.println(i);  
6.   }  
7.  }  
8.  public static void main(String args[]){  
9.   TestCallRun2 t1=new TestCallRun2();  
10.   TestCallRun2 t2=new TestCallRun2();  
11.    
12.   t1.run();  
13.   t2.run();  
14.  }  
15. } 

Output:1
       2
       3
       4
       5
       1
       2
       3
       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.