Split a number into its digits with Haskell

Question

Given an arbitrary number, how can I process each digit of the number individually?

Edit I've added a basic example of the kind of thing Foo might do.

For example, in C# I might do something like this:

static void Main(string[] args) {     int number = 1234567890;     string numberAsString = number.ToString();      foreach(char x in numberAsString)     {         string y = x.ToString();         int z = int.Parse(y);         Foo(z);     } }  void Foo(int n) {     Console.WriteLine(n*n); } 

Answer 1

Have you heard of div and mod?

You'll probably want to reverse the list of numbers if you want to treat the most significant digit first. Converting the number into a string is an impaired way of doing things.

135 `div` 10 = 13 135 `mod` 10 = 5 

Generalize into a function:

digs :: Integral x => x -> [x] digs 0 = [] digs x = digs (x `div` 10) ++ [x `mod` 10] 

Or in reverse:

digs :: Integral x => x -> [x] digs 0 = [] digs x = x `mod` 10 : digs (x `div` 10) 

This treats 0 as having no digits. A simple wrapper function can deal with that special case if you want to.

Note that this solution does not work for negative numbers (the input x must be integral, i.e. a whole number).

Answer 2

digits :: Integer -> [Int] digits = map (read . (:[])) . show 

or you can return it into []:

digits :: Integer -> [Int] digits = map (read . return) . show 

or, with Data.Char.digitToInt:

digits :: Integer -> [Int] digits = map digitToInt . show 

the same as Daniel's really, but point free and uses Int, because a digit shouldn't really exceed maxBound :: Int.