Getting ArrayIndexOutOfBoundsException after a Certain INPUT value

Question

for the Given below code after int Input Value of 46348 i am getting ArrayIndexOutOfBoundsException. I am given Condition in for loop that keeps the array limits. But somehow i am getting this exception and i unable to figure it out. And my requirement is find all primenumbers below given number.

 Scanner sc = new Scanner(System.in);
    int n= sc.nextInt();
    int[] arr= new int[n+1];
            for(int i=2;i<=n;i++)
            {
                if(arr[i]==0)
                {
                    for(j=i;j*i<=n;j++)
                        arr[j*i]=1; // Here i am getting Exception
                }
            }

Input:

46349

Output:

java.lang.ArrayIndexOutOfBoundsException: -2146737495

502802

Thanks.,

buc · Accepted Answer · 2012-06-26 07:31:12Z

3

You have encountered an arithmetic overflow.

In Java, int data type is a 32-bit signed integer, which means it can have values between -2147483648 and 2147483647.

On this line:

for(j=i;j*i<=n;j++)

If i is 46349 then j becomes 46349, too. If you multiply 46349 by 46349, you get 2148229801, which is greater than 2147483647, so the integer overflows and becomes -2146737495. Naturally, it is less than 46349, so the check in the for-loop passes. But you cannot index an array with a negative value in Java, that's why you get the ArrayIndexOutOfBoundsException.

Range check your input value for n < 46340, or if it really needs to work with n = 46349 input, switch to long data type, which will work up to n = 3037000499.

edited Jun 26, 2012 at 7:31

answered Jun 26, 2012 at 7:16

buc

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Comments

Affe · Accepted Answer · 2012-06-26 07:16:46Z

0

46349 * 46349 is too big to use as the index of a Java Array. The index is just a 32-bit signed integer, so has a maximum value of 2,147,483,648.

It passes the < n check because it overflows and comes back negative, so it is in fact less than n, but a negative number is not a legal array index.

answered Jun 26, 2012 at 7:16

Affe

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