[Haskell-beginners] Couldn't match expected type `(a0 -> [a0]) -> [a0]', with actual type `[a0]'

[Sumit Sahrawat, Maths & Computing, IIT (BHU)]

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 <r.wobben at home.nl> 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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