Will this ever result in a stack overflow error?
- by David
Will incrementing the instance variables of an object ever lead to a stack overflow error?
For example:
This method (java) will cause a stack overflow error:
class StackOverflow {
public static void StackOverflow (int x)
{
System.out.println (x) ;
StackOverflow(x+1) ;
}
public static void main (String[]arg) { StackOverflow (0) ;
}
but will this?: (..... is a gap that i've put in to shorten the code. its long enough as it is.)
import java.util.*;
class Dice
{
String name ;
int x ;
int[] sum ;
....
public Dice (String name)
{
this.name = name ;
this.x = 0 ;
this.sum = new int[7] ;
}
....
public static void main (String[] arg)
{
Dice a1 = new Dice ("a1") ;
for (int i = 0; i<6000000; i++)
{
a1.roll () ;
printDice(a1) ;
}
}
....
public void roll ()
{
this.x = randNum(1, this.sum.length) ;
this.sum[x] ++ ;
}
public static int randNum (int a, int b)
{
Random random = new Random() ;
int c = (b-a) ;
int randomNumber = ((random.nextInt(c)) + a) ;
return randomNumber ;
}
public static void printDice (Dice Dice)
{
System.out.println (Dice.name) ;
System.out.println ("value: "+Dice.x) ;
printValues (Dice) ;
}
public static void printValues (Dice Dice)
{
for (int i = 0; i<Dice.sum.length; i++)
System.out.println ("#of "+i+"'s: "+Dice.sum[i]) ;
}
}
The above doesn't currently cause a stack overflow error but could i get it too if i changed this line in main: for (int i = 0; i<6000000; i++) so that instead of 6 million something sufficiently high were there?
