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
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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