Determining the amount of digits, and storing the values in array

[CooKieLord]

Situation: I want to break down an integer number into a bunch of digits, and store the value in an array.
E.g. The user enters 1 337.

The program will break it down to 1, 3, 3, 7 and store each digit in an array such that:
integerArray[0] = 1
integerArray[1] = 3
integerArray[2] = 3
integerArray[3] = 7

I understand that I can use a combination of divisions and modulo in order to determine this, but the problem I have is that I need to determine the number of digits. This is what I have, but there's are logic errors and I can't figure them out.

Java:

public class test {    public static void main  (String[] args) {       int[] array;  //Array to store broken down number       int bigNumber; //User-inout       int digit = 1; //Number of digits       int i =0, digitMod, expo; //dunno? O_O i is for array index, digitMod and expo are test variables       bigNumber = ITI1120.readInt(); //ITI1120 is the class that they gave me to read numbers             //Block 1 - Determine the amount of digits (Need help here)       for (int d = 0; bigNumber % digit != digit; digit += 10) {          digit = d; }          array = new int[digit];       //Block 2 - Breaking down the digits       i = digit -1;       expo = 1;       while (i >= 0) {          digitMod = Math.pow (10, expo);          array[i] = (bigNumber / digitMod) % 10;          expo++;          i--; //         digitMod += 10; };       //Block 3 - Printing the results       for (int x = 0; x < digit; x++)          System.out.print(array[x] + "\n");          } // end main }//end class

[HeavenAgain]

try making the number into a string first, then its automatically in an array
just have to convert

[Euphoracle]

Well, you could do it that way, or you could do it the easy way and use a string.

Java:

public clas test {         public static void main(String[] args)         {                 int[] digits;                 int input = 0;                 int digitCount = 0;                 String quickCount = ""; // Hehehehe                 input = ITI1120.readInt(); // I would prefer to use a long, but meh                 quickCount = input + "";                 digitCount = quickCount.length(); // Oh, how I love doing things the easy way some times                 digits = new int[digitCount];                                 for(int i = 0; i < digitCount; i++)                 {                         digits[i] = Integer.parseInt(quickCount.charAt(i)); // :p                 }                                 for(int i = 0; i < digitCount; i++)                         System.out.println(digits[i]);         } }

Of course, if you're not allowed to do that, this won't help much Sorry for any syntax errors if any, I only have access to notepad at the moment and no JDK.

[richcash]

If you take the log (base 10) of a number and round down, the result will be 1 less than the number of digits in that number.

Remember, Math.log() in java.lang.Math is the natural log, so you have to use the base changing formula :

code:

Math.log(num)/Math.log(10)

will give you the log (base 10) of a number.

Do you know how to get the number of digits now? (Note, you'll have to make adjustments for 0 and negative numbers if you choose to include them).

As you said, you can use division and modulus once you know the number of digits.

[richcash]

In case you're having trouble :

Java:

int num = 9846711; int numOfDigits = (int) (Math.log (num) / Math.log(10)) + 1;

Post if you need help with storing the individual digits.

[wtd]

code:

class MySpecialInteger {     private int value;     public MySpecialInteger() {         this(0);     }     public MySpecialInteger(int value) {         this.value = value;     }     public int intValue() {         return value;     }     public int[] digits() {         Integer[] temp = digitsHelper(value, new java.util.ArrayDeque<Integer>());         int index = 0;         int[] digits = new int[temp.length];                 for (Integer v : temp) {                         digits[index++] = v;         }                 return digits;     }     private static Integer[] digitsHelper(       int value,       java.util.ArrayDeque<Integer> accumulator) {         if (value > 0) {             accumulator.addFirst(value % 10);             return digitsHelper(value / 10, accumulator);         }         else {             return (Integer[])accumulator.toArray();         }     } }

Recursion FTW!

[richcash]

I would rather write your digits() method like this.

Java:

public int[] digits() {         int[] digits = new int[(int) (Math.log(value) / Math.log(10)) + 1];         for (int i = 0; i < digits.length; i ++) {                 digits[i] = (int) (value / Math.pow (10, i)) % 10;         }         return digits; }

But I guess thinking recursively about it is cool too.

[wtd]

You will also notice that my digitsHelper method is tail-recursive.

[CooKieLord]

Thank for all the help guys, and sorry I'm a little late in replying.

I'm mainly looking at the way richcash showed me. It's simpler for me, because I don't quite understand how to get recursion to work yet.

Now, the only problem I have is storing the broken down values in the array. With the way you're shown to me, it would place the units in [0], the 10s in [1], the 100s in [2], etc. I need the highest digit, let's say 1000s in [0], the 100s in [1], the 10s in [2] and the 1s in [3]

(I'm talking about array subscripts right? or is it called index?)

Don't tell me how to do it. I'll edit this post if I can't figure it out. I just wanted to let you guys know that I read the posts. Thanks again!

[Euphoracle]

Reverse the array:

Java:

int[] reverse(int[] source) {         int[] reversed = new int[source.length];         // loop through, replace source into reversed backwards        // code removed, sorry, didn't read that last bit         return reversed; }

[CooKieLord]

Well this is what I did

Java:

public class test {    public static void main  (String[] args) {       int[] array;       int bigNumber;       double digit;       int digint;       bigNumber = ITI1120.readInt();       digit = Math.floor(Math.log(bigNumber) / Math.log(10)) + 1;       array = new int[(int) digit];       int placement = (int) array.length;       for (int index=0; index < array.length; index ++) {          array[index] = (int) ((bigNumber / Math.pow (10, (placement-1))) % 10);          placement--; };       for (int x = 0; x < array.length; x++)          System.out.print(array[x] + "\n");    } // end main }//end class

I had two variables, one for the array and the other for the modulo. Is that what you meant by "reversing"?

[Euphoracle]

Your method does work, yes.