Question

I’m trying to write a function in haskell that reverses lists recursively. I wrote a helper function that takes the original list and an empty list then transfers elements from the first one to the other in a LIFO pattern.

Here’s what I have:

myreverse :: [a] -> [a]
myreverse list = myflip list []

myflip :: [a] -> [a] -> [a]
myflip list1 newList
    | null list1        = newList
    | otherwise         = myflip (tail list1) ((head list1) : newList)

I know there’s a built-in function that does it for me but the requirement is that I only use head, tail, elem and null (No pattern matching either). So my question is: is there a better solution where I only have one function, myreverse, that consumes only one list? (That meets the requirements above, of course)

Thanks!

Answer 1

So for anyone who might be interested, this is what I ended up writing:

myreverse :: [a] -> [a]
myreverse list = myflip list []
    where myflip list newList = if null list then newList
                                else myflip (tail list) ((head list):newList)

Thanks everyone for your comments and suggestions.