calling main method inside main in java

You can. And it is very much legal.
Also, it doesn't necessarily give a StackOverFlowError:

public static void main(String args[]){ int randomNum = (int)(Math.random()*10); if(randomNum >= 8) System.exit(0); System.out.print("Heyhey!"); main(new String[]{"Anakin", "Ahsoka", "Obi-Wan"}); } 

Just saying.

You will get StackOverFlowError. If you call endless recursive calls/deep function recursions.

Thrown when a stack overflow occurs because an application recurses too deeply.

You need to pass String array like

main(new String[] {"a","b","c"}); 

Kugathasan is right.

Yes it does give StackOverflow Exception. I tested it right now in Eclipse.

class CallingMain { public static void main(String[] args) { main(new String[] {"a","b","c"}); } } 

I have one suggestion, I think the best way to eliminate the confusion is to try coding and running it. Helps with lots of doubts.

You can simply pass the argument like main(new String[1]) - this line will call main method of your program. But if you pass "new String[-1]" this will give NegativeArraySizeException. Here I'mm giving an example of Nested Inner Class where I am calling the Main method:

class Outer { static class NestedInner { public static void main(String[] args) { System.out.println("inside main method of Nested class"); }//inner main }//NestedInner public static void main(String[] args) { NestedInner.main(new String[1] );//**calling of Main method** System.out.println("Hello World!"); }//outer main }//Outer 

Output:- inside main method of Nested class

Hello World!