CODEWITHSUNDEEP
javaarraysjava.util.scanner
user8360486
user8360486

Reputation: 29

Why is the scanner function not working?

So this code takes the value of n and returns a list of divisors as well as the total number of divisors. If I remove the Scanner declaration and its assignment to int n and simply give int n a value, the code runs perfectly.

However, as it is, it returns this: 

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 0
    at Program.main(Program.java:25)

I have no idea what the problem is.

import java.util.Scanner;
public class Program{
    static int n;
    static int x = 1;
    static int [] arr = new int[n];
    static int q = 0;
    static int g = 0;
    static int p = 1;
    static int count;


    public static void main(String[] args){

         Scanner scan = new Scanner(System.in);
         int n = scan.nextInt();

         while(x <= n){

            arr [q] = p; //assigns value to each array index
            g = n%arr[q]; // stores value of remainder
            q++; 
            p++;
            x++;
            if (g == 0){ //counts and displays each time remainder  = 0
                count++;
                System.out.println(q);
            }

        }

        System.out.println(count + " Divisors");


}
}

Upvotes: 1

Views: 203

Answers (3)

user7627726
user7627726

Reputation:

The size of arr is declared when n still holds no value(before the size is inputted). Do this:

import java.util.Scanner;

public class Program {
    static int n;
    static int x = 1;
    static int [] arr; //no size set
    //...
    //Other variables
    //...

    public static void main(String[] args){
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        arr = new int[n]; //Now the size is set to the inputted number.
        while(x <= n) {
            //...
            //Other code to find divisors
            //...
        }
    }
}

You need to name the size of arr after you input the n, otherwise the size is set to 0, causing the ArrayIndexOutOfBoundsException.

This line:

arr[q] = p;

Is what actually caused the error. arr[q] couldn't hold a value, because there was no arr[q]. the array had no size, so it couldn't hold any members.

Upvotes: 2

developer_hatch
developer_hatch

Reputation: 16224

Change your while condition from

while(x <= n)

to

while(x < n)

< means strict less than, so you start from 1, and not get the out of bounds

Edit:

Also as @CodingNinja said, you have to change the and define a int value to the size of the array, by default is 0:

public static void main(String[] args){
     static int n;
     Scanner scan = new Scanner(System.in);
     int n = scan.nextInt();


     static int [] arr = new int[n];

     while(x <= n){

        arr [q] = p; //assigns value to each array index
        g = n%arr[q]; // stores value of remainder
        q++; 
        p++;
        x++;
        if (g == 0){ //counts and displays each time remainder  = 0
            count++;
            System.out.println(q);
        }

    }

    System.out.println(count + " Divisors");


  }

Upvotes: 0

geokavel
geokavel

Reputation: 619

static int n;
...
static int [] arr = new int[n];

You do not give n a value, so it defaults to 0. Therefore, you initialize arr as an array of length 0. That is why you get Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 0. It's because your array is size 0, so even index 0 is out of bounds of the array.

If you don't know n until you read from the Scanner you should change your code to:

public static void main(String[] args){

     Scanner scan = new Scanner(System.in);
     int n = scan.nextInt();
     int [] arr = new int[n];
 ...
 }

Upvotes: 1

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