Is it possible to nest guards in Haskell?

Question

Haskell newbie here, trying to write code to parse math expressions. Code:

isDigit :: Char -> Bool
isDigit c = c >= \'0\' && c <= \'9\'

pars         


        

Answer 1

Recent GHC now has MultiWayIf:

{-# LANGUAGE MultiWayIf #-}

parseNumber :: String -> Maybe (String, String)
parseNumber [] = Just ("", "")
parseNumber (h:ls)
  | isDigit h = if
    | p == Nothing -> Just ([h], ls)
    | otherwise    -> Just (h:fst d, snd d)
  | h == '.'  = if
    | p == Nothing             -> Nothing
    | not ('.' `elem` (snd d)) -> Just (h:(fst d), snd d)
  | otherwise = Nothing
  where p@(~(Just d)) = parseNumber ls

But this is better written slightly differently anyhow, without the partiality.

{-# LANGUAGE MultiWayIf #-}

parseNumber :: String -> Maybe (String, String)
parseNumber [] = Just ("", "")
parseNumber (h:ls)
  | isDigit h = if
    | Nothing <- p -> Just ([h], ls) -- PatternGuards, on by default
    | Just d  <- p -> Just (h:fst d, snd d)
  | h == '.'  = if
    | Nothing <- p                         -> Nothing
    | Just d  <- p, not ('.' `elem` snd d) -> Just (h:(fst d), snd d)
  | otherwise = Nothing
  where p = parseNumber ls

and you may as well use maybe.

parseNumber :: String -> Maybe (String, String)
parseNumber "" = Just ("", "")
parseNumber (h:hs)
  | isDigit h = maybe (Just ([h], hs)) (\(num, rest') -> Just (h:num, rest')) rest
  | h == '.'  = maybe Nothing (\(num, rest') -> if '.' `elem` num then Nothing
                                                else Just (h:num, rest')
                              ) rest -- This logic is a bit wonky; it doesn't really work
  | otherwise = Nothing
  where rest = parseNumber hs