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.
snippet
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
 }
}
snippet
Output:running...
MainThreadStack Problem if you direct call run() method
snippet
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
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.
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