I think you can check full source code at its web site to see what you
missed.
My thought: did you make Parser as a instance of Monad?
-Haisheng
On Mon, Jul 18, 2011 at 5:20 PM, anyzhen
well,i am reading the book :<*Programming in haskell *> *chapter 8* *functional parsers * this chapter try teaching an parser technology,but i have some problem about *do* keyword
--some brief info of types : type Parser a = String -> [(a,String)] item :: Parser Char --item "abc" = [('a',"bc")] return' :: a -> Parser a --return' 'a' "foo" = [('a',"foo")]
--below code complier notice... sorry format p2 :: Parser (Char,Char) p2 = do x <- item item y <- item return' (x,y) *"No instance for (Monad ((-)) String)) * * arising from a execute statement ** ** ** ** ** * * Possible fix : add an instance declaration for **(Monad ((-)) String))** * *In a stmt of a 'do' expression : y <- item** ** ** ** ** * *In the expression:** ** ** ** ** ** ** ** * * do {** x <- item;** ** ** ** ** ** ** ** * * item;** ** ** ** ** ** ** ** ** ** * * y <- item;** ** ** ** ** ** ** ** * * return' (x,y)}** ** ** ** ** ** ** * * In an equation for 'p2':** ** ** ** ** ** ** ** * * p2 ** ** ** ** ** ** ** ** ** * * = do { ** x <- item;** ** ** ** ** ** ** * * item;** ** ** ** ** ** ** ** * * y <- item;** ** ** ** ** ** ** * * ....**}"** ** ** ** ** ** ** ** * * * *thanks for any help * *jiangzhen mail:jiangzhen3s@qq.com*
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