[Haskell-beginners] expression parser

[Daniel Fischer]

Am Dienstag 22 Dezember 2009 21:48:11 schrieb John Moore:
> Hi,
>    Could some one point out what I'm doing wrong below. This is a parser
> which takes an arithmetic expression built up from single digits. Turns
> 3*4+2 into (3*4)+2 etc.It giving me the last expression in a do statement
> or parse error on ->
>
> (+++) :: Parser a -> Parser a -> Parser a
> expr :: term( '+' expr |"")
> term -> factor('*' term |"")

What's happened here?
A part of your grammar seems to have slipped into the code.
>
> expr :: Parser Int
> expr  = do t <- term
>    do char '+'
>       e <- expr
>       return (t + e)
>     +++ return t
>
> term :: Parser Int
> term = do f <- factor
>       do char '*'
>          t <- term
>       return (f * t)
>    +++ return f
>
> factor :: Parser Int
> factor = do d <- digit
>              return (digitToInt d)
>   +++ do char '('
>          e <- expr
>       char ')'
>       return e
>
> eval :: String -> Int
> eval xs = fst(head(parse expr xs))


Maybe it's just the mail programme, but what arrived here had wrong indentation.
In expr, the inner do-block isn't aligned with "t <- term".
In term, the "return (f * t)" isn't aligned with "char '*'" and "t <- term" and the whole 
inner dop-block isn't aligned with "f <- factor".
In factor,  "return (digitToInt d)" isn't aligned with "d <- digit" and in the second 
block, "char ')'" and "return e" aren't aligned with "char'('" and "e <- expr".

Properly aligned and with an implementation of (+++) instead of grammar productions, it 
should work.
>
>
> John