Integer to Integer Array C#

A simple solution using LINQ

 int[] result = yourInt.ToString().Select(o=> Convert.ToInt32(o)).ToArray()

I believe this will be better than converting back and forth. As opposed to JBSnorro´s answer I reverse after converting to an array and therefore avoid IEnumerable´s which I think will contribute to a little bit faster code. This method work for non negative numbers, so 0 will return new int[1] { 0 }.

If it should work for negative numbers, you could do a n = Math.Abs(n) but I don't think that makes sense.

Furthermore, if it should be more performant, I could create the final array to begin with by making a binary-search like combination of if-statements to determine the number of digits.

public static int[] digitArr(int n)
{
    if (n == 0) return new int[1] { 0 };

    var digits = new List<int>();

    for (; n != 0; n /= 10)
        digits.Add(n % 10);

    var arr = digits.ToArray();
    Array.Reverse(arr);
    return arr;
}

Update 2018:

public static int numDigits(int n) {
    if (n < 0) {
        n = (n == Int32.MinValue) ? Int32.MaxValue : -n;
    }
    if (n < 10) return 1;
    if (n < 100) return 2;
    if (n < 1000) return 3;
    if (n < 10000) return 4;
    if (n < 100000) return 5;
    if (n < 1000000) return 6;
    if (n < 10000000) return 7;
    if (n < 100000000) return 8;
    if (n < 1000000000) return 9;
    return 10;
}

public static int[] digitArr2(int n)
{
    var result = new int[numDigits(n)];
    for (int i = result.Length - 1; i >= 0; i--) {
        result[i] = n % 10;
        n /= 10;
    }
    return result;
}

int[] outarry = Array.ConvertAll(num.ToString().ToArray(), x=>(int)x);

but if you want to convert it to 1,2,3,4,5:

int[] outarry = Array.ConvertAll(num.ToString().ToArray(), x=>(int)x - 48);