How to Turn an Int into an Array of Ints of Each Digit

Convert an integer to an array of digits

The immediate problem is due to you using <= temp.length() instead of < temp.length(). However, you can achieve this a lot more simply. Even if you use the string approach, you can use:

String temp = Integer.toString(guess);
int[] newGuess = new int[temp.length()];
for (int i = 0; i < temp.length(); i++)
{
    newGuess[i] = temp.charAt(i) - '0';
}

You need to make the same change to use < newGuess.length() when printing out the content too - otherwise for an array of length 4 (which has valid indexes 0, 1, 2, 3) you'll try to use newGuess[4]. The vast majority of for loops I write use < in the condition, rather than <=.

How to convert an integer into an array of digits

I'd go with

var arr = n.toString(10).replace(/\D/g, '0').split('').map(Number);

You can omit the replace if you are sure that n has no decimals.

Converting an integer to an array of digits

Off the top of my head:

int i = 123;
var digits = i.ToString().Select(t=>int.Parse(t.ToString())).ToArray();

Is there an easy way to turn an int into an array of ints of each digit?

public Stack<int> NumbersIn(int value)
{
    if (value == 0) return new Stack<int>();

    var numbers = NumbersIn(value / 10);

    numbers.Push(value % 10);

    return numbers;
}

var numbers = NumbersIn(987654321).ToArray();

Alternative without recursion:

public int[] NumbersIn(int value)
{
    var numbers = new Stack<int>();

    for(; value > 0; value /= 10)
        numbers.Push(value % 10);

    return numbers.ToArray();
}

convert an integer number into an array

This would work for numbers >= 0

#include <math.h>

char * convertNumberIntoArray(unsigned int number) {
    int length = (int)floor(log10((float)number)) + 1;
    char * arr = new char[length];
    int i = 0;
    do {
        arr[i] = number % 10;
        number /= 10;
        i++;
    } while (number != 0);
    return arr;
}

EDIT: Just a little bit more C style but more cryptic.

#include <math.h>

char * convertNumberIntoArray(unsigned int number) {
    unsigned int length = (int)(log10((float)number)) + 1;
    char * arr = (char *) malloc(length * sizeof(char)), * curr = arr;
    do {
        *curr++ = number % 10;
        number /= 10;
    } while (number != 0);
    return arr;
}

How to put an integer to an array of digits

First calculate no of digits

int count = 0;
int n = number;

while (n != 0)
{
    n /= 10;
    cout++;
}

Now intialize the array and assign the size:

if(count!=0){
   int numberArray[count];

   count = 0;    
   n = number;

   while (n != 0){
       numberArray[count] = n % 10;
       n /= 10;
       count++;
   }
}

Converting integer into array of digits

Just using something like this:

int n = 544; // your number (this value will Change so you might want a copy)
int i = 0; // the array index
char a[256]; // the array

while (n) { // loop till there's nothing left
    a[i++] = n % 10; // assign the last digit
    n /= 10; // "right shift" the number
}

Note that this will result in returning the numbers in reverse order. This can easily be changed by modifying the initial value of i as well as the increment/decrement based on how you'd like to determine to length of the value.

(Brett Hale) I hope the poster doesn't mind, but I thought I'd add a code snippet I use for this case, since it's not easy to correctly determine the number of decimal digits prior to conversion:

{
    char *df = a, *dr = a + i - 1;
    int j = i >> 1;

    while (j--)
    {
        char di = *df, dj = *dr;
        *df++ = dj, *dr-- = di; /* (exchange) */
    }
}

How do I separate an integer into separate digits in an array in JavaScript?

Why not just do this?

var n =  123456789;
var digits = (""+n).split("");