I have written enough code in the IO monad to fairly well understand how
the 'do' and 'bind' forms work. But I've never seen monadic code that was
NOT in the IO monad. Your explanation tells me where to go do more study.
Thank you.If I go read the definition of bind in the List monad, it will
probably become clear. In the original code, the List monad is completely
invisible to my untrained eye, and I was confused.
On Sun, Jul 7, 2019 at 12:13 PM Francesco Ariis
Hello Terry
On Sun, Jul 07, 2019 at 11:24:47AM -0400, Terry Phelps wrote:
I found this code on the net somewhere. It compiles and works properly:
import qualified Data.ByteString as BS import Text.Printf (printf) toHex :: BS.ByteString -> String toHex bytes = do hex <- BS.unpack bytes printf "%02x" hex
I cannot understand the 'do' notation is required, because it seems to be a pure function. I guess there's a monad hiding somewhere that my newbie mind can't see.
`toHex` is pure (non IO), but it has an /effect/. In this case, it takes advantage of the list monad to achieve non-determinism.
Specifically, since
unpack :: ByteString -> [Word8]
printf (which in our case has signature (`String -> Char`) gets called on each of those [Word8]. The result will obviously be [Char], which `String` is an alias of.
So, I rewrote the code to remove the 'do stuff':
[...] toHex :: BS.ByteString -> String toHex bytes = printf "02x" (BS.unpack bytes)
A do-less version still is /monadic/, hence it will have >>= or >> or similar somewhere. This works:
toHex2 :: BS.ByteString -> String toHex2 bytes = BS.unpack bytes >>= printf "%02x"
and follows the reasoning above (feed every every Word8 to `printf "%02x"`).
Does this answer your questions? -F
_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners