Part 1 Chapter 7 Exercise 6, part 3.

Author: PiotrJustyna

part1_chapter7_exercise6/Main.hs

@@ -18,6 +18,12 @@ main = do
     putStrLn "chop8' [1, 2, 3, 4, 5, 6, 7, 8, 9]"
     putStrLn . show $ chop8' [1, 2, 3, 4, 5, 6, 7, 8, 9]
 
+    putStrLn "map (+1) [1, 2, 3]"
+    putStrLn . show $ map (+1) [1, 2, 3]
+
+    putStrLn "map' (+1) [1, 2, 3]"
+    putStrLn . show $ map' (+1) [1, 2, 3]
+
 unfold :: (a -> Bool) -> (a -> a) -> (a -> a) -> a -> [a]
 unfold predicate headFunction tailFunction x
     | predicate x = []
@@ -31,4 +37,53 @@ chop8 [] = []
 chop8 bits = (take 8 bits) : (chop8 (drop 8 bits))
 
 chop8' :: [Bit] -> [[Bit]]
-chop8' = unfold null (take 8) (drop 8)
+chop8' = unfold null (take 8) (drop 8)
+
+-- This one was quite confusing to me, so I thought I would leave some comments to complement the implementation.
+-- The author asks readers to implement "map f". "map" signature is:
+--
+-- map :: (a -> b) -> [a] -> [b]
+--
+-- which translates to (for lists, that is): every item in list [a] gets mapped from type a to type b and then
+-- returned as [b].
+--
+-- Now with that in mind, does it really makes sense to apply "unfold" here? Is the function not supposed to build a list
+-- using the starting value and the functions provided? If not, I'm not sure what really is required. If so, though,
+-- the type signature defined by the regular "map" is going to make:
+-- the implementation unnecessarily complicated
+-- map and map' are not going to be interchangable.
+--
+-- My initial thought was to implement function:
+--
+-- map' :: (a -> b) -> [a] -> [b]
+--
+-- with the signature identical to the regular "map" (just a different implementation) so that both can be used interchangeably.
+-- It took me a while before I realized that it is probably more tricky than it should be.
+-- This is when I started looking for other solutions, just to get an idea how people interpret the task.
+-- What I found mademe even more confused. Examples:
+--
+-- 1. https://github.com/evturn's implementaion: https://github.com/evturn/programming-in-haskell/blob/master/07-higher-order-functions/07.9-exercises.hs
+-- 
+-- map' p f xs = unfold p f f xs
+--
+-- the type here is:
+--
+-- map' :: (a -> Bool) -> (a -> a) -> a -> [a]
+--
+-- So as you can see, it doesn't match the original "map"'s type and the input argument (second last "a") gets "unfolded" into [a].
+--
+-- 2. Another example I found is the one by https://github.com/RoccoMathijn: https://github.com/RoccoMathijn/programming-in-haskell/blob/master/chapter07.hs
+--
+-- map' :: (a -> b) -> [a] -> [b]
+-- map' f = unfold (null) (f.head) (tail)
+--
+-- But this again is not intercangeable with the original "map" as b is simply [a], so the result is really [[a]].
+--
+-- After more time than I'd like to admit, I finally figured out the implementation which is interchangeable with the original "map".
+-- Terribly inefficient, but I believe it answers the exercise's question precisely.
+
+map' :: (a -> b) -> [a] -> [b]
+map' f = concat . unfold null (listify . f . head) (tail)
+
+listify :: a -> [a]
+listify x = [x]