4 ms·

The Haskell 98 Report takes two lines: foldr f z [] = z foldr f z (x:xs) = f x (foldr f z xs) Clojure does it like this (note the overloading -

• eru 17y ago
  To be true, it's customary that you also add a line for the types in Haskell. Though not necessary.
• jdeseno 17y ago
  Couldn't we do this? r' f (x:xs) | xs == [] = (f x):[] | otherwise = (f x):(r' f xs) or: [(f a) | a <- myList] - another edit - r' f (x:xs) | xs == [] = x | otherwise = f x (r' f xs)
  • mbrubeck 17y ago
    That's map, not reduce.
    • deleted 17y ago
      [deleted]
      • mbrubeck 17y ago
        That's filter, not reduce. :)
        • jganetsk 17y ago
          He's right with the second one. r' f (x:xs) | xs == [] = x | otherwise = f x (r' f xs) r' :: (Eq t) => (t -> t -> t) -> [t] -> t That's definitely not filter. That's definitely reduce. (I'm equating reduce with foldr1 or foldl1). Now, because he used the == operator, that adds the (Eq t) constraint. We can rewrite his code as the following... r' f (x:xs) = case xs of {[] -> x; _ -> f x (r' f xs)} r' :: (t -> t -> t) -> [t] -> t Try actually running it on some code. r' (+) [1,2,3,4] evaluates to 10
          • mbrubeck 17y ago
            The "filter" version was deleted and the correct version added, after I wrote my comment.