Integer to Integer Array C# [duplicate]

A simple solution using LINQ

int[] result = yourInt.ToString().Select(o=> Convert.ToInt32(o) - 48 ).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);