Re: why does this work?

Daniel Pitts <>
Wed, 08 Aug 2012 10:38:27 -0700
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 way.

Hopefully this helps.

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