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Implementing Runnable

 
Greenhorn
Posts: 4
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this code outputs a sequence of 5, 10, 15.. but if i change the value at i%5 to i%6 it will wtill output 5 but in a sequence of +6 like 5, 11, 17, 23, 29 ...
my problem is how come it started with 5 as the output and why, since 5 is not intialized to int i anywhere in the code and why does it follow that sequence, even when i change the value of i%5 to i%4, it still follows the same sequence in i%6. why?
from k&b
page 6

import java.lang.Runnable;

class Wombat implements Runnable {
private int i;
private int x;
public synchronized void run() {
if (i%4 != 0) { i++; }
for(int x=0; x<5; x++, i++)
{ if (x > 1) Thread.yield(); }
System.out.print(i + " ");
}
public static void main(String[] args) {
Wombat n = new Wombat();
for(int x=100; x>0; --x) { new Thread(n).start(); }
} }
 
Ranch Hand
Posts: 113
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Hi
By Default i is getting default value 0
therefore
(i%4) or (i%5) or (i% 6) will give value 0
((i%4)!=0) will give false therefore I will not get incremented and cursor move to for loop
after for loop value of i and x is 5
therefore (i%5) will give zero and i will not increment
but (i%4) and (i%6) will give remainder 1 and 5 so i will get increment
 
Remy coolcoin
Greenhorn
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thank you Gala for the reply.
Is my pleasure to hear from you.

My Regards
 
Don't get me started about those stupid light bulbs.
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