Exception error in arrays

Question

My code is intended to take in an array of ints and return an array of ints with a length of twice the original array's length, minus 2.

The returned array should have values 1/3 and 2/3 between any given two values in the original array.

For example input array:

{400, 500, 600}

will return:

{400,433,466,500,533,566,600}

The code I have is as follows:

public static void main(String[] args){
            System.out.println("Problem 9 tests");
            int[] arr7={300, 400, 500};
            System.out.println(highDef(arr7));
            System.out.println(" ");

public static int[] highDef(int[] original) {
        int[] newarr = new int[original.length*3-2];
        newarr[0]=original[0];
        int count=0;
        while (count!=newarr.length) {
                int increment=(original[count+1]-original[count])/3;
                newarr[count+1]=original[count]+increment;
                newarr[count+2]=original[count]+(count*increment);
                count+=1;
        }
        return newarr;

while (count!=newarr.length -2 ) add "-2" to the while condition, you can't get newarr[count+1] when count == newarr.length — jdabrowski
– jdabrowski, Commented Mar 7, 2016 at 22:26
Look within your loop - imagine when count == newarr.length - 2... but then you're accessing newarr[count+2] in the body of the loop. (I'd strongly advise you to use for loops for this sort of thing, by the way...) — Jon Skeet
– Jon Skeet, Commented Mar 7, 2016 at 22:26
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 3 at Lab2a.highDef(Lab2a.java:186) at Lab2a.main(Lab2a.java:26) — Ashley Gold
– Ashley Gold, Commented Mar 7, 2016 at 22:27

jdabrowski · Accepted Answer · 2016-03-07 23:19:30Z

1

Haven't looked good enough the first time :) Here you go:

public static void main(String[] args) {
    System.out.println("Problem 9 tests");
    int[] arr7 = {300, 400, 500};
    System.out.println(Arrays.toString(highDef(arr7)));
    System.out.println(" ");
}

public static int[] highDef(int[] original) {
    int[] newarr = new int[original.length * 3 - 2];
    newarr[original.length * 3 - 3] = original[original.length-1];

    for(int i=0; i<original.length-1; i++){
        newarr[i*3] = original[i];
        int increment = (original[i+1] - original[i])/3;
        newarr[i*3+1] = original[i] + increment;
        newarr[i*3+2] = original[i] + increment*2;
    }

    return newarr;
}

edited Mar 7, 2016 at 23:19

answered Mar 7, 2016 at 22:27

jdabrowski

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2 Comments

Ashley Gold Over a year ago

Still receiving the same out of bounds exception :(

Ashley Gold Over a year ago

This was very helpful in helping me understand what the exception means, but for some reason adding -3 to the while condition still did not fix the error. I'm not sure why

Rana · Accepted Answer · 2016-03-07 22:38:22Z

0

With array out of bound exceptions, try to think of the last case.

In your example:

original.length = 3

newarr.length = (original.length * 3) - 2 = 7

while (count!=newarr.length) // this means: while count does not equal to 7. Therefore *Count* value can go upto 6
// imagine the last case, count = 6
newarr[count+2]=original[count]+(count*increment); // the problem is here
// count(6) + 2 = 8, which is out of bounds

same issue with this line: int increment=(original[count+1]-original[count])/3;

count will go upto 6 but original[count+1] = original[7] which is out of bound as well

answered Mar 7, 2016 at 22:38

Rana

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