Re: [Haskell-cafe] recursion problem part 2

Dmitry Olshansky schreef op 6-2-2015 om 16:25:

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

Thanks that worked, I have now to figure out why I see [0] on 0 and no zero when it reaches it end-point. isDigits 0 n< 10 so [0] IsDigits 1 n < 0 not true n < 10 not true. toDigits 1 ++ [0] toDigits 1 + [0] n< 0 not true n < 10 true so [1] ++ [0] [1] ++ [0] is [1.0] where the output is [1] So somehow I do not understand fully how this works . Roelof