Confusing output from infinite recursion within try-catch

Question

Consider the following code.

public class Action {
private static int i=1;
public static void main(String[] args)          


        

Answer 1

axtavt Answer is very complete but I'd like to add this:

As you may know the stack is used to store the memory of variables, based on that you cannot create new variables when you reach the limit, it is true that System.out.println will need some stack resources

787     public void More ...println(Object x) {
788         String s = String.valueOf(x);
789         synchronized (this) {
790             print(s);
791             newLine();
792         }
793     }

Then after calling the print, the error does not allow you to even call the newLine, it breaks again right on the print. Based on that you can make sure that's the case by changing your code like this:

public class Action {
    static int i = 1;

    public static void main(String[] args) {
        try {
            System.out.print(i + "\n");
            i++;
            main(args);
        } catch (StackOverflowError e) {
            System.out.print(i + " SO " + "\n");
            i++;
            main(args);
        }
    }
}

Now you will not ask the stack to handle the new lines, you will use the constant "\n" and you may add some debugging to the exception printing line and your output will not have multiple values in the same line:

10553
10553 SO
10553 SO
10554
10554 SO
10554 SO
10555
10556
10557
10558

And it will keep broken until get some resources to allocate new data and pass to the next i value.