Index out of bound in java

Question

Why it index out of bound?

I nap a picture in ---> PICTURE

Code:

Scanner sc = new Scanner(System.in);
int a=0,b=0,n=0;
int[][] sum;
int q = sc.nextInt();

for(int i=0;i<q;i++){

    a = sc.nextInt();
    b = sc.nextInt();
    n = sc.nextInt();
    sum = new int[i][n];

    sum[i][0] = a +  (int)Math.pow(2,0) * b;

focus on this

    sum = new int[i][n];
    sum[i][0] = a +  (int)Math.pow(2,0) * b;

Input:

1
5 3 5

Exception:

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: Index 0 out of bounds for length 0
at Main.main(Main.java:19)

Please check this post to understand more about `ArrayIndexOutOfBoundsException ` https://stackoverflow.com/questions/5554734/what-causes-a-java-lang-arrayindexoutofboundsexception-and-how-do-i-prevent-it — Nagaraju Chitimilla, Sep 20 '21 at 06:20
Please don't post code as image. See https://meta.stackoverflow.com/questions/285551/why-not-upload-images-of-code-errors-when-asking-a-question why. — Reporter, Sep 20 '21 at 06:25
Here: `sum = new int[i][n];`, when `i = 0` in your first loop iteration, you create an array of size 0. You then try to access the array of size 0 at index 0 here `sum[i][0] = a + (int)Math.pow(2,0) * b;`, which is why you get the exception. See the linked duplicate. — maloomeister, Sep 20 '21 at 06:26
@NagarajuChitimilla Okay i will read rule immediately , Thanks for edit my post ;_; to be good for read. — Green Goblan, Sep 20 '21 at 07:02

score 0 · Answer 1 · edited Sep 20 '21 at 07:00

0

For your first iteration, i = 0 and you initialized an int array (int[0][5]) which is an empty array. size is 0, 5.

And at the same time you are accessing an element at the position 0. In order to get an element at position 0 then the size of the array should be > 0.

edited Sep 20 '21 at 07:00

user16320675

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answered Sep 20 '21 at 06:35

Hassan Liasu

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Okay, I got it . Thanks a lot for answer my first post ;_; – Green Goblan Sep 20 '21 at 07:08

Index out of bound in java

1 Answers1