Recursive functions that reverse a list in Haskell

Question

I've been tasked to create a recursive function that takes in a list and gives the reverse of the list.

I have been able to create the a function named rev1 which does this:

rev1 :: [a] -> [a] 
rev1 [] = [] 
rev1 (x:xs) = reverse xs ++ [x] 

But I have been asked to create a another function 'rev2' which should use an additional argument serving as an accumulator.

Could somebody please help me compose the rev2 function.

Thanks in advance, Sam

Answer 1

First of all, your rev1 should be this instead:

rev1 :: [a] -> [a] 
rev1    []  = [] 
rev1 (x:xs) = rev1 xs ++ [x]

The point of rev2 would be to achieve tail recursion by means of passing the intermediate result inside an accumulator argument:

rev2 :: [a] -> [a] -> [a]
-- if applied to an empty list, just return the result
-- so far, i.e. whatever is inside the accumulator:
rev2 acc    []  =         acc
-- otherwise, take the head of the list and append it to
-- the accumulator, and carry on with the rest of the list
rev2 acc (x:xs) = rev2 (x:acc) xs

this, however, obviously has the downside of exposing the acc argument to users of rev2 , so a typical approach is hiding the accumulator based implementation behind a façade that looks exactly like rev1 :

rev2 :: [a] -> [a]
rev2 xs = go [] xs where
  go :: [a] -> [a] -> [a]  -- signature included for clarity
  go acc    []  =       acc
  go acc (x:xs) = go (x:acc) xs

Answer 2

Let me start things off:

rev2 :: [a] -> [a]
rev2 xs = rev2' xs ???

rev2' :: [a] -> [a] -> [a]
rev2' [] ys = ???
rev2' (x : xs) ys = ???