[Haskell-beginners] recursion problem.

[Roelof Wobben]

Hello,

I have figured out how I can make from 123   [1,2,3]

I did it this way :

-- | convert a number to a array in pieces where a negative number will 
be a empty array.
toDigits :: Integer -> [Integer]
toDigits n
    | n <= 0 = []
    | otherwise = toDigits (n `div` 10) ++ [n `mod` 10]


but now when I do toDigits 0 , I see [] as output where I was expected [0]

But when I do  toDigits 100 I see the right output [ 1,0,0]  which 
surprises me because I tought that with the first 0 there will be a []

Or is the n the second time [0,0]  and the thirth time [0] So it will be 
like this :

toDigits 100

to Digits [1] ++ [ 0,0]
toDigits [1] ++ [0] ++  [0]

which leads to [1,0,0]

Roelof