Weird couldn't match type error

Question

I have simple one line function:

revRange :: (Char,Char) -> [Char]
revRange t = unfoldr (\b -> if b == (pred (fst t)) then Nothing else Just (b, pred b)) (snd t)

It works well:

*Main Data.List> revRange ('a', 'f')
"fedcba"

Then I want to extract unfoldr lambda as unique function:

revRange :: (Char,Char) -> [Char]
revRange t = unfoldr fun t
fun t = (\b -> if b == (pred (fst t)) then Nothing else Just (b, pred b)) (snd t)

Now, I have a weird error:

Couldn't match type `Char' with `(Char, Char)'
Expected type: (Char, Char) -> Maybe (Char, (Char, Char))
  Actual type: (Char, Char) -> Maybe (Char, Char)
In the first argument of `unfoldr', namely `fun'
In the expression: unfoldr fun t

Answer 1

First, format your code:

revRange :: (Char,Char) -> [Char]
revRange t = unfoldr fun t
fun t = (\b -> if b == (pred (fst t)) 
               then Nothing 
               else Just (b, pred b)) (snd t)

Next, double check the type of unfoldr using Hoogle:

unfoldr :: (b -> Maybe (a, b)) -> b -> [a]

Next, add a type signature to fun so that GHC will tell you what the problem is. According to the type of unfoldr, fun should have the type:

b ~ Char
a ~ Char

fun :: Char -> Maybe (Char, Char)

since you are unfoldr in the original example with snd t.

I usually like to verify small pieces, so we can remove the definition of foo and just use the type signature:

revRange :: (Char,Char) -> [Char]
revRange t = unfoldr fun t

fun:: Char -> Maybe (Char, Char)
fun b = error ""

GHC complains that t has type (Char, Char) but fun expects type Char. You've called unfoldr fun t instead of unfoldr fun (snd t) as in the original example. Move that bit from fun into revRange:

revRange :: (Char,Char) -> [Char]
revRange t = unfoldr fun (snd t)

fun:: Char -> Maybe (Char, Char)
fun b = error ""

Next, add in the definition of fun again. We can remove the lambda and put b as a normal argument to fun:

fun:: Char -> Maybe (Char, Char)
fun t b = if b == (pred (fst t)) 
          then Nothing 
          else Just (b, pred b)

Immediately we see another glaring problems: fun takes two arguments but the signature says it should only take one!

Since t is a constant in the original lambda, we can solve this problem by partially applying fun in revRange, so the final answer is:

revRange :: (Char,Char) -> [Char]
revRange t = unfoldr (fun t) (snd t)

fun:: (Char, Char) -> Char -> Maybe (Char, Char)
fun t b = if b == (pred (fst t)) 
          then Nothing 
          else Just (b, pred b)

---

To address your comment, you'd like to write

revRange :: (Char,Char) -> [Char]
revRange = unfoldr fun2

Using the same approach as above, in the signature of unfoldr we need b ~ (Char,Char) and a ~ Char. So we'd like fun2 to have the type

fun2 :: ((Char,Char) -> Maybe (Char, (Char, Char)))

I'll leave the definition of fun2 as an exercise. As a hint, I suggest adopting the convention that the first part of the pair is constant and holds fst t.