haskell - parsing n hex digits using attoparsec -

okay need parse n digits of hex , having problem cant stop standard attoparsec hex parser hexadecimal.

my first idea this:

nhex n = take n *> hexadecimal doesnt work because takes off 4 digits parses rest of string xd

next idea works this:

hex :: (num a, eq a) => int -> parser hex n = fst . head . readhex <$> count n (satisfy ishexdigit) 

but problem code in attoparsec library warns against returning lists of chars speed concerns , hex parser base of whole program

next idea try better speed this:

parsefragments :: (bits a, integral a) => int -> parser parsefragments n =       fourchars <- b.take n       let hexdigits = parseonly hexadecimal fourchars       case hexdigits of                 left err -> fail err               right x  -> return x 

but feels terrible hack using parseonly. there fast way more idiomatic?

data.attoparsec.bytestring.char8.hexadecimal implemented as:

hexadecimal :: (integral a, bits a) => parser hexadecimal = b8.foldl' step 0 `fmap` i.takewhile1 ishexdigit       ishexdigit w = (w >= 48 && w <= 57) ||                    (w >= 97 && w <= 102) ||                    (w >= 65 && w <= 70)     step w | w >= 48 && w <= 57  = (a `shiftl` 4) .|. fromintegral (w - 48)              | w >= 97             = (a `shiftl` 4) .|. fromintegral (w - 87)              | otherwise           = (a `shiftl` 4) .|. fromintegral (w - 55) 

you can use pretty same, except need insepct result of take, of characters might not valid hexadecimal characters. use (maybe -> word8 -> maybe a) put both in same function, simplicity, i've used functions above:

fixedhexadecimal :: (integral a, bits a) => int -> parser fixedhexadecimal n =     bytes <- a.take n     if b8.all ishexdigit bytes        b8.foldl' step 0 bytes       else fail "fixedhexadecimal"    ishexdigit = -- see above         step       = -- see above