如何将一串逗号分隔值解析为haskell中的字符串列表？

Question

so if I have a string "(this, is, a, story, all, about, how)" into a list of the words inside it ["this", "is", "a", "story", "all", "about", "how"] as an instance of ReadP String? I've tried a bunch of different ways, one of which being this:

parseStr :: ReadP String
parseStr = do
  skipSpaces
  n <- munch1 isAlphaOrDigit
  skipComma
  return $ n

which parses all values but the last. I thought if I combined it with this parse:

parseLast :: ReadP String
parseLast = do
  skipSpaces
  n <- munch1 isAlphaOrDigit
  return $ n

as

parseLet = (many parseStr) +++ parseLast

but that didn't work either. Any tips?

edit: more definitions

isAlphaOrDigit :: Char -> Bool 
isAlphaOrDigit a = (isDigit a) || (isAlpha a) 
comma = satisfy (','==)
skipComma = const () <$> some comma

Answer 1

The parser a +++ b sends the entire input string to a and the entire input string to b , producing all the results that either parser produced. You instead want a parser that sends the first part of the string to a and the second part to b , then lets you combine the results. Try this instead:

parseLet = liftA2 (\ss s -> ss ++ [s]) (many parseStr) parseLast

Many parser libraries also offer a manySepBy combinator (perhaps with a slightly different name) for this exact use case; you might consider looking through the ReadP library for such a thing.