Re: why does this work?
To: dkoleary
From: "Daniel Pitts" <daniel.pitts@1:261/38.remove-qhs-this>
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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