Re: [Haskell-beginners] Question about code I found

It's more correct to say the List Monad Is a Monad instance who is a List But people look at you funny In my case Funnier than usual On Sun., Jul. 7, 2019, 9:47 a.m. Terry Phelps, wrote:

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 wrote:

...

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

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