Question

isTogether' :: String -> Bool
isTogether' (x:xs) = isTogether (head xs) (head (tail xs))

For the above code, I want to go through every character in the string. I am not allowed to use recursion.

Answer 1

isTogether' (x:xs) = isTogether (head xs) (head (tail xs))

If I’ve got it right, you are interested in getting consequential char pairs from some string. So, for example, for abcd you need to test (a,b), (b,c), (c,d) with some (Char,Char) -> Bool or Char -> Char -> Bool function.

Zip could be helpful here:

> let x = "abcd"
> let pairs = zip x (tail x)
it :: [(Char, Char)]

And for some f :: Char -> Char -> Bool function we can get uncurry f :: (Char, Char) -> Bool.

And then it’s easy to get [Bool] value of results with map (uncurry f) pairs :: [Bool].