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From: "Daniel Pitts" <daniel.pitts@1:261/38.remove-qhs-this>
Subject: Re: why does this work?
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Date: Wed, 08 Aug 2012 19:04:28 GMT
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  To: dkoleary
From: Daniel Pitts <newsgroup.nospam@virtualinfinity.net>

On 8/8/12 10:30 AM, dkoleary wrote:
> Hi;
>
> 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 XCopy.java
> $ 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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