Split a number into its digits with Haskell

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).

Using the same technique used in your post, you can do:

digits :: Integer -> [Int]
digits n = map (\x -> read [x] :: Int) (show n)

See it in action:

Prelude> digits 123
[1,2,3]

Does that help?

You could also just reuse digits from Hackage.

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.