Am 03.07.2012 15:16, schrieb Robert Heumüller:
I am sorry, I must have hit the wrong hotkey. My code looks like this:
type Parser a = String -> [(a, String)]
This type synonym is unsuitable for a Monad instance. Better would be: newtype Parser a = Parser (String -> [(a, String)]) but that would require to change your code below.
result :: a -> Parser a result v = \inp -> [(v, inp)]
zero :: Parser a zero = \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 -> concat [f v inp' | (v, inp') <- p inp]
Such a definition (without signature and adapted to the newtype) should be used within a Monad instance. "instance Monad Parser where ..."
sat :: (Char -> Bool) -> Parser Char sat p = do x <- item if p x then result x else zero
Without proper Monad instance you should not use "do". Instead you could expand it yourself manually to: sat p = item Main.>>= \x -> if p x then result x else zero Note the clash between your ">>=" function (Main.>>=) and the one from the Prelude! HTH Christian P.S. there is a "instance Monad ((->) r))" in Control.Monad.Instances but that does not fit your parser type, too.
I beleive I understand how this code is meant to work, but when I run it in ghci I get the follwing error-message:
parser.hs:21:13: No instance for (Monad ((->) String)) arising from a do statement Possible fix: add an instance declaration for (Monad ((->) String)) In a stmt of a 'do' block: x <- item In the expression: do { x <- item; if p x then result x else zero } In an equation for `sat': sat p = do { x <- item; if p x then result x else zero }
parser.hs:22:18: Couldn't match expected type `Char' with actual type `[(Char, String)]' In the first argument of `p', namely `x' In the expression: p x In a stmt of a 'do' block: if p x then result x else zero Failed, modules loaded: none.
Sadly I have no idea how to fix this :(
Thanks again :)