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Head First Java - Chapter 4 puzzle question

 
Greenhorn
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I'm trying to understand the pool puzzle problem on page 91 and hoping that someone can give me a hand. I can't figure out how the integer (ivar) gets any value assigned to it other than 0 and I can't understand how it works with 0 as the value. Below is a copy of the code.

public class Puzzle4 {
public static void main(String[] args) {
Puzzle4b[] obs = new Puzzle4b[6];
int y = 1;
int x = 0;
int result = 0;
while (x < 6) {
obs[x] = new Puzzle4b();
obs[x].ivar = y;
y = y * 10;
x = x + 1;
}
x = 6;
while ( x > 0 ) {
x = x - 1;
result = result + obs[x].doStuff(x);
}
System.out.println("result " + result);
}
}
class Puzzle4b {
int ivar;
public int doStuff(int factor) {
if (ivar > 100) {
return ivar * factor;
}else{
return ivar * ( 5 - factor );
}
}
}

p.s. This is my third head first book and I love them all.
 
author and iconoclast
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Hi,

Welcome to JavaRanch!

Just looking at the Puzzle4b class by itself, you don't see the variable being assigned to anywhere. But look at Puzzle4, and you see

obs[x] = new Puzzle4b();
obs[x].ivar = y;

these two lines create a Puzzle4b instance, and set the value of ivar in that instance. Later, when obs[x].doStuff(x) is called, the Puzzle4b instance in obs[x] still has that nonzero value of ivar.
 
William Wyatt
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Thank you very much, especially for the quick response. It seems so obvious now, don't know why I didn't see it before. Thanks again.
 
William Wyatt
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One more quick question, in this section:

while ( x > 0 ) {
x = x - 1;
result = result + obs[x].doStuff(x);
}
System.out.println("result " + result);
}
}

Will the while loop run when x is equal to 0?
 
Ernest Friedman-Hill
author and iconoclast
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Nope.
 
Ranch Hand
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it won't but x will receive the value 0 when you enter the loop at x=1, because of the line x=x-1.
 
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