Here's a solution:
lastDigit x = mod x 10
remainingDigits x = (x - lastDigit x) `div` 10
listDigits x
  | x < 10 = [x]
  | otherwise = (listDigits $ remainingDigits x) ++ [lastDigit x]

Here's a faster one:
listDigits2 x
  | x < 10 = [x]
  | otherwise = (listDigits $ remainingDigits) ++ [lastDigit]
  where lastDigit = mod x 10
        remainingDigits = (x - lastDigit) `div` 10


On Fri, Feb 6, 2015 at 8:07 AM, Roelof Wobben <r.wobben@home.nl> wrote:
Jerzy Karczmarczuk schreef op 6-2-2015 om 16:57:

Le 06/02/2015 16:48, Roelof Wobben a écrit :
Im only sure that somehow I take the wrong turn somewhere because I cannot figure out why I do not get the right answer.

I think I have the value of n wrong.
Did you read my answer entirely? Do so, find the red fat line

You take n=1, and you write:  n<10 not True.
You don't need to be a specialist on recursion, to see the error.

Jerzy


oke,

Another try :

toDigits :: Integer -> [Integer]
toDigits n
   | n < 0 = []
   | n < 10 = [n]
   | otherwise =  toDigits (n `div` 10) ++ [n `mod` 10]

 isDigits 1

n < 0 not true.
n < 10 true so [1]

 IsDigits 12

n <  0 not true
n < 10 not true
 toDigits  1 + [2]

toDigits 1 + [2]

n < 0 not true
n < 10 true [1] ++ [2]

1 ++2 = [1,2]

I think I understand it finnaly.


Roelof

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