Try this: toDigits' :: Integral a => [a] -> a -> [a] toDigits' acc 0 = acc toDigits' acc n = toDigits' (n `mod` 10 : acc) (n `div` 10) If the accumulator in a recursive function doesn't accumulate anything during some recursive call, but is just passed on unchanged, then you might as well make it a free variable. Or eliminate it entirely if it's unused. You can now use a functional programming idiom, and assign the function toDigits to the function toDigits' [], like so: toDigits :: Integral a => a -> [a] toDigits = toDigits' [] On Sat, May 23, 2015 at 6:59 AM, Sumit Sahrawat, Maths & Computing, IIT (BHU) wrote:

The error lies here:

toDigits' acc number = ((number `mod` 10 ): acc) (toDigits' (number `mod` 10))

It should instead be

toDigits' acc number = (number `mod` 10) : (toDigits' acc (number `mod` 10))

My suggestion would be to look at it like a fold.

toDigits' :: Integral a => a -> a -> [a] toDigits' acc 0 = [acc] toDigits' acc n = n `mod` 10 : toDigits' acc (n `div` 10)

Now this gives the digits in the reverse order, so in toDigits, you can reverse it.

A good exercise would now be to re-write this as a fold. Graham Hutton has a good paper about it. [1]

The best way would be to directly convert the number to a string using show, but that's not the point of the exercise.

[1]: https://www.cs.nott.ac.uk/~gmh/fold.pdf

On 23 May 2015 at 12:28, Roelof Wobben wrote:

...

Hello,

For some reasons my file's are corrupted. I had repair them with success except this one.

Here is the code :

toDigits number | number <= 0 = [] | otherwise = toDigits' [] number where toDigits' acc 0 = acc toDigits' acc number = ((number `mod` 10 ): acc) (toDigits' (number `mod` 10))

main = print $ toDigits 123

and here is the error:

Couldn't match expected type `(a0 -> [a0]) -> [a0]' with actual type `[a0]' The function `(number `mod` 10) : acc' is applied to one argument, but its type `[a0]' has none In the expression: ((number `mod` 10) : acc) (toDigits' (number `mod` 10)) In an equation for toDigits': toDigits' acc number = ((number `mod` 10) : acc) (toDigits' (number `mod` 10))

Roelof

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-- Regards

Sumit Sahrawat

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