
(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
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

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,
(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.

That's a valid point. come to think of it, is there any use case where Void
appears on a positive position?
On Sun, Jan 13, 2019 at 1:15 AM Isaac Elliott
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.
-- Javran (Fang) Cheng

Functions like `forever` can put Void in positive position as a proof that
the value is never produced.
On Sun, Jan 13, 2019 at 11:53 AM Javran Cheng
That's a valid point. come to think of it, is there any use case where Void appears on a positive position?
On Sun, Jan 13, 2019 at 1:15 AM Isaac Elliott
wrote: 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.
-- 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.
participants (3)
-
Isaac Elliott
-
Javran Cheng
-
Michael Snoyman