Re: [Haskell-cafe] Fwd: Void is not Monoid?

If you had an `RWST r Void s m a`, then you would be able to produce an `m (a, s, Void)`, which is `absurd`. On Sun, 13 Jan. 2019, 6:43 pm Javran Cheng, wrote:

(forgot to reply all, sorry)

Hi Will,

Thanks for the reply!

...

A monoid has an identity element, and void does not.

now I feel dull never thought about that.

...

How would you write return with void as the writer? You can accomplish what you want with the free monoid over Void - i.e. [Void], which is isomorphic to unit. So unit seems like the right choice.

Unit does work fine, but I figure using Void is an interesting idea, as I can make sure that no one can use the "W" part of my RWST.

Javran

On Sun, Jan 13, 2019 at 12:23 AM William Yager wrote:

...

On Sun, Jan 13, 2019 at 4:00 PM Javran Cheng wrote:

...

Hi Cafe,

I'm wondering why Data.Void does not have a Monoid instance, or, what would be the problem if we do "mempty = absurd mempty"?

This diverges, does it not?

A monoid has an identity element, and void does not.

...

Long story: I was using a monad with some transformers, then I realize I can collapse State and Reader into RWST with Void being Writer output. (well, I could have just used Unit but I wanna give Void a try...) I know beforehand that Void is Semigroup but is a bit surprise it doesn't have Monoid instance.

How would you write return with void as the writer?

You can accomplish what you want with the free monoid over Void - i.e. [Void], which is isomorphic to unit. So unit seems like the right choice.

--Will

-- Javran (Fang) Cheng

-- Javran (Fang) Cheng _______________________________________________ Haskell-Cafe mailing list To (un)subscribe, modify options or view archives go to: http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe Only members subscribed via the mailman list are allowed to post.