-- this should work too
parseHeader3 :: BS.ByteString -> Maybe (Int, Int)
--note accurate type signature, which helps us use Maybe failure-monad,
--although losing your separate error messages
parseHeader3 bs = do
(x, rest) <- BS.readInt $ BS.dropWhile (not . isDigit) bs
(y, _) <- BS.readInt $ BS.dropWhile (not . isDigit) rest
return (x, y)
--or to be clearer without syntactic sugar, that is
parseHeader3 bs =
(BS.readInt $ BS.dropWhile (not . isDigit) bs)
>>= \(x, rest) ->
(BS.readInt $ BS.dropWhile (not . isDigit) rest)
>>= \(y, _) ->
return (x, y)
Isaac
Paulo J. Matos wrote:
On Dec 23, 2007 12:32 PM, Paulo J. Matos <[EMAIL PROTECTED]> wrote:
Hello all,
It is either too difficult to get two integers of a bytestring, in
which case something should be done to ease the process or I should
learn much more Haskell. I guess the latter is the correct guess.
I have a bytestring containing two naturals. I was to get them as
efficiently as possible. Here's my code:
Just tried a better one, what do you think of this:
parseHeader2 :: BS.ByteString -> (Int, Int)
parseHeader2 bs =
case (BS.readInt $ BS.dropWhile (not . isDigit) bs) of
Nothing -> error "Couldn't find first natural."
Just (x, rest) ->
case (BS.readInt $ BS.dropWhile (not . isDigit) rest) of
Nothing -> error "Couldn't find second natural."
Just (y, _) -> (x, y)
parseHeader :: BS.ByteString -> (Int, Int)
parseHeader bs =
let first = BS.readInt $ BS.dropWhile (not . isDigit) bs
in
if(isNothing first)
then
error "Couldn't find first natural."
else
let second = BS.readInt $ BS.dropWhile (not . isDigit) $
snd $ fromJust first
in
if(isNothing second)
then
error "Couldn't find second natural."
else
(fst $ fromJust first, fst $ fromJust second)
This seems to work:
parseHeader $ BS.pack "hello 252 359"
(252,359)
Is there a better way?
Cheers,
--
Paulo Jorge Matos - pocm at soton.ac.uk
http://www.personal.soton.ac.uk/pocm
PhD Student @ ECS
University of Southampton, UK
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