[Haskell-beginners] variable parser

[Daniel Fischer]

Am Samstag 09 Januar 2010 18:45:23 schrieb John Moore:
> Hi,
>     I just want to add a new parser for variables but this doesn't seem
> to work can anybody tell me what I'm doing wrong?
>
>
>
> module ExprSyn where
> import Text.ParserCombinators.Parsec
> import Text.ParserCombinators.Parsec.Expr
> runExpr :: String -> IO ()
> runExpr str = case  runParser expr () " "  str of
>    Left err -> putStrLn "Error:" >> print err
>    Right val -> print val
> demo1 = runExpr  "2+5"
> data Expr = Mul Expr Expr
>
>          | Div Expr Expr
>          | Add Expr Expr
>          | Sub Expr Expr
>          | Val Float
>   |
>   |Let String Expr Expr
>   | Var String
>
>  deriving (Eq,Show)
> expr :: Parser Expr
> expr = variable <|> buildExpressionParser table factor
>    <?> "expression"

variable succeeds on the start of the input "2". Then the second parser 
isn't tried at all. Try reversing the alternatives:

expr = buildExpressionParser table factor <|> variable <?> "expression"

Perhaps even better and less surprising, don't let variable parse numbers, 
so

expr = buildExpressionParser table factor <|> variable <|> number <?> 
"expression"

> table :: [[Operator Char st Expr]]
> table = [[op "*" Mul AssocLeft, op "/" Div AssocLeft]
>        ,[op "+" Add AssocLeft, op "-" Sub AssocLeft]
>        ]
>  where
>    op s f assoc
>       = Infix (do{ string s; return f}) assoc
> factor :: Parser Expr
> factor = do{ char '('
>           ; x <- expr
>           ; char ')'
>           ; return x
>           }
>      <|> number
>      <|> variable
>      <?> "simple expression"
> variable :: Parser Expr
> variable = do { b <- letter
>        ; do { cs <- word
>        ;return (c:cs)

Wrong type, should be
   return (Var (c:cs))
resp.
   return (Var c])

>   <|> return [c]
>   <|> number
>        }
>
> number :: Parser Expr
> number = do{ ds <- many1 digit
>           ; return (Val $ read ds)
>           }
>
>
>  John