I mean powers of *ten* :) On Dec 8, 2007 10:48 PM, Philip Weaver wrote:

Well, you're choosing to parse each digit of your integer as a separate integer, so if you want to combine them after reading you'll need to multiply by powers of two. Or, you can just read in all the digits in one 'read' command, like this:

parseInt :: String -> (Expr, String) parseInt xs = let (digits, rest) = span isDigit in (EInt (read digits), rest)

where 'span' is defined in the Prelude. Hope this helps!

- Phil

On Dec 8, 2007 10:03 PM, Ryan Bloor wrote:

...

hi

The code below does almost what I want but not quite! It outputs...parseInt "12444a" gives... [(EInt 1,"2444a"),(EInt 2,"444a"),(EInt 4,"44a"),(EInt 4,"4a"),(EInt 4,"a")]

What I want is: [(EInt 12444, "a")]

data Expr = EInt {vInt :: Int} -- integer values | EBool {vBool :: Bool} -- boolean values

parseInt :: Parser parseInt (x:xs) | (isDigit x && xs /= []) = [(EInt (read [x]),xs)] ++ parseInt xs | isDigit x && xs == [] = [(EInt (read [x]),[])] | otherwise = []

Thanks

Ryan

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