In Java , why cant the main method be called inside the main method ? Like a recursive function.

Question

See the following code
It gives an error (Why?) :

abstract class Parent
{
public static void main (String[] arg)
{
System.out.println("\n This is the Parent class");
Sub_class_1 obj1 = new Sub_class_1();
Sub_class_2 obj2 = new Sub_class_2();
obj1.message();
obj2.message();
for ( int i=1; i<=3; i++ )
{
main();
/* Since main method is static , hence no object . Also , main method is in the class Parent , and we are calling main method inside the class , so only function is name is enough (no class required ). */
}

}

}

class Sub_class_1 extends Parent
{
public void message()
{
System.out.println("\n This is first subclass");
}
}

class Sub_class_2 extends Parent
{
public void message()
{
System.out.println("\n This is second subclass");
}
}

Answer 1

You can:
public class MainInMain{
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[]{"Recursion", "Gone", "Dayve"});
}}
Pass the main with a String...so as it is done ;)

P.S. You can apply this to your code as well!

Answer 2

There's another way called NestedInner:

class Outer {
static class NestedInner {
public static void main(String[] args) {
System.out.println("Inside the main");
}//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

Answer 3

You couldn't because main() was not a method. The signature does not match anything. You need to pass an array of type string as the arguments.
Ex/
main(arg);
main is just a method, it works like any others.

Answer 4

http://stackoverflow.com/questions/21992659/calling-main-method-inside-main-in-java
http://stackoverflow.com/questions/22624983/call-main-recursively
it seems like you can do that. look the links.