[Haskell-beginners] Question about code I found

[KC]

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, <tgphelps50 at gmail.com> 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 <fa-ml at ariis.it> 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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