On Mon, May 9, 2011 at 5:07 PM, Eric Rasmussen
Hi everyone,
I am relatively new to Haskell and Parsec, and I couldn't find any articles on parsing numbers in the following format:
Positive: $115.33 Negative: ($1,323.42)
I'm working on the parser for practical purposes (to convert a 3rd-party generated, most unhelpful format into one I can use), and I'd really appreciate any insight into a better way to do this, or if there are any built-in functions/established libraries that would be better suited to the task. My code below works, but doesn't seem terribly efficient.
Why do you think it inefficient? Is it slow? I don't have any substantial suggestions, but from a style perspective: * I would question the use of IEEE binary-floating-point number types for currency. Haskell ships with a fixed-point decimal library, but I don't know how fast it is: http://hackage.haskell.org/packages/archive/base/latest/doc/html/Data-Fixed.... * You can re-use the 'positive' parser in the 'negative' one:
negAmount = do char '(' a <- posAmount char ')' return (negate a)
* In the 'currencyAmount' declaration, I'm not sure that you need the 'try' before the 'negAmount', but I don't know what the rest of your grammar looks like.
Thanks! Eric
------------------------------------------------- {- parses positive and negative dollar amounts -}
integer :: CharParser st Integer integer = PT.integer lexer
float :: CharParser st Double float = PT.float lexer
currencyAmount = try negAmount <|> posAmount
negAmount = do char '(' char '$' a <- currency char ')' return (negate a)
posAmount = do char '$' a <- currency return a
currency = do parts <- many floatOrSep let result = combine orderedParts where combine = sumWithFactor 1 orderedParts = reverse parts return result
floatOrSep = try float <|> beforeSep
beforeSep = do a <- integer char ',' return (fromIntegral a :: Double)
sumWithFactor n [] = 0 sumWithFactor n (x:xs) = n * x + next where next = sumWithFactor (n*1000) xs
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