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

[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, <javran.c at gmail.com> 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 <will.yager at gmail.com>
> wrote:
>
>> On Sun, Jan 13, 2019 at 4:00 PM Javran Cheng <javran.c at gmail.com> 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
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