Non-exhaustive patterns in function haskell

Question

I think I am missing the case where there's a one element list but I can't find a way to write it can someone help me?

getBoard :: [String] -> [String]
getBoard (h:t) | isLOL h = h: getBoard (t)
               | otherwise = []


isLOL :: String -> Bool
isLOL [ ] = True
isLOL (h:t) | h>='a' && h<='z' || h >='A' && h<='Z' = isLOL t
            | otherwise = False

Answer 1

getBoard [] = []

is the line you want. Like this:

getBoard :: [String] -> [String]
getBoard [] = []
getBoard (h:t) | isLOL h = h: getBoard (t)
               | otherwise = []


isLOL :: String -> Bool
isLOL [] = True
isLOL (h:t) | h>='a' && h<='z' || h >='A' && h<='Z' = isLOL t
            | otherwise = False