Hello I've been reading Programming in Haskell, and I'm trying to go through the parser examples in chapter. However, I'm getting a type error when using the "do" notation. Here's the code I'm trying to load in ghci, which is copied from the book: import Prelude hiding ((>>=), return) import Char type Parser a = String -> [(a, String)] return :: a -> Parser a return v = \inp -> [(v, inp)] failure :: Parser a failure = \inp -> [] item :: Parser Char item = \inp -> case inp of [] -> [] (x:xs) -> [(x, xs)] parse :: Parser a -> String -> [(a, String)] parse p inp = p inp (>>=) :: Parser a -> (a -> Parser b) -> Parser b p >>= f = \inp -> case parse p inp of [] -> [] [(v, out)] -> parse (f v) out p :: Parser (Char, Char) p = do x <- item item y <- item return (x, y) The problem is in the definition of "p": Couldn't match expected type `Char' against inferred type `[(Char, String)]' In the expression: x In the first argument of `return', namely `(x, y)' In the expression: return (x, y) Now if I define p as p :: Parser (Char, Char) p = item >>= \x -> item >>= \_ -> item >>= \y -> return (x, y) it works fine, so I'm wondering what else I need to do for the "do" notation to work. Thanks in advance, Andre