What if we call run() method directly

by Online Tutorials Library July 22, 2022

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

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

FileName: TestCallRun1.java

  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   }  }  

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Output:

running...

Problem if you direct call run() method

FileName: TestCallRun2.java

    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();   }  }  

Test it Now

Output:

1  2  3  4  1  2  3  4

As we 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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