Re: why does this work?

"Daniel Pitts" <daniel.pitts@1:261/38.remove-k2r-this>
Thu, 09 Aug 2012 18:44:36 GMT
  To: dkoleary
From: "Daniel Pitts" <daniel.pitts@1:261/38.remove-qhs-this>

  To: dkoleary
From: Daniel Pitts <>

On 8/8/12 10:30 AM, dkoleary wrote:


New java programmer. So new, in fact, that I'm still working my way through

the O'Reilly Head First Java book. One of the end of chapter questions
involves identifying if a sample class will compile and what to do to make it
compile if it won't.

The sample class from chapter 4 is:

class XCopy
{ public static void main(String[] args)
    { int orig = 42;
       XCopy x = new XCopy();
       int y = x.go(orig);
       System.out.println(orig + " " + y);

    int go(int arg)
    { return arg * 2; }

The book says that it'll compile and run, displaying "42 84" and, sure

enough, it does:

$ javac
$ java XCopy
42 84

How come that isn't recursive? XCopy.main() instantiates a new XCopy.

Shouldn't that new XCopy instance also instantiate a new XCopy? new XCopy()
creates a new instance of the XCopy class, which executes Constructors. main
isn't executed again because of this. main is only automatically executed by
the JVM on start-up, once.

I was figuring this would run until the XCopy.go function tried returning a

number that wouldn't fit in int anymore... That's obviously not the case, but I
don't know why.
int can fit any number in the range [-2^31, 2^31). 42 and 84 are both within
that range. It is possible that if you pass in a number with a large enough
magnitude, you will end up with an overflow. In Java (and many 2s-compliment
integer systems), overflow will simply throw-away the upper bits, and you will
have what is called "wrap-around". This actually makes many of the basic
operations easier, because signed numbers and unsigned numbers behave the same

Hopefully this helps.

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