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
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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