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!
- PhilOn Dec 8, 2007 10:03 PM, Ryan Bloor <ryanbloor@hotmail.com> 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
Get closer to the jungle. I'm a Celebrity Get Me Out Of Here!
_______________________________________________
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe