Java - Storing the output of a for loop into an array

Question

I'm using a for loop to create the first 400 multiples of 13, and I'm trying to to store these numbers into an array. The specific issue is on 5th line. I understand that this code is making the programme write to the first element of the array, which is causing the issue. Is there any way I can store the numbers sequentially?

    public static void main(String[] args) {
    int[] thirteenMultiples = new int[400];
    for (int dex = 1; dex < thirteenMultiples.length; dex ++) {
        int multiples = 13 * dex; 
        thirteenMultiples[0] = multiples;
        System.out.println(Arrays.toString(thirteenMultiples));

Use dex, since it's incremented each loop: thirteenMultiples[dex] = multiples. — saagarjha
– saagarjha, Commented Jan 25, 2018 at 17:49

bkis · Accepted Answer · 2018-01-25 18:00:25Z

6

Array indices start at 0, so change int dex = 1 to int dex = 0. Also, you should use your counting variable dex to write to the right array index:

public static void main(String[] args) {
    int[] thirteenMultiples = new int[400];
    for (int dex = 0; dex < thirteenMultiples.length; dex ++) {
        int multiples = 13 * dex; 
        thirteenMultiples[dex] = multiples;
        System.out.println(Arrays.toString(thirteenMultiples));
    }
}

BTW: Arrays.toString(thirteenMultiples) is quite an expensive operation to do on every iteration (try to code this method yourself and you'll see what i mean). Maybe you should just print the current value of thirteenMultiples[dex] and print you array once the loop has finished. I assume you're just testing and trying stuff for now, but i think it's good to keep such things in mind from the beginning ;)

edited Jan 25, 2018 at 18:00

answered Jan 25, 2018 at 17:49

bkis

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Neeraj Yadav · Accepted Answer · 2018-01-25 17:50:54Z

4

thirteenMultiples[dex] in place of thirteenMultiple[0], because dex is equal to the index each time for loop runs. For ex - for dex =1 you store multiple at [1], then it increases to 2 then it becomes [2] and you store the next multiple at 2. Hence it stores each new value at new index.

Also start dex from 0 as array starts from 0 index.

answered Jan 25, 2018 at 17:50

Neeraj Yadav

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Ahmed Kamal · Accepted Answer · 2018-01-25 17:52:16Z

0

I guess in this case, it is better to use a List rather than an array. Your code will look cleaner

public static void main(String[] args) {
    List<Integer> thirteenMultiples = new ArrayList<Integer>;
    for (int dex = 0; dex < 400; dex ++) {
        thirteenMultiples.add(13 * dex)
    }
    System.out.println(thirteenMultiples);

}

answered Jan 25, 2018 at 17:52

Ahmed Kamal

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1 Comment

Chris Dunning Over a year ago

Thank you for the alternate perspective. I'll look into this.

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