Give MonadTrans a QuantifiedConstraints superclass

[Zemyla]

return is lift . return. Upon reflection, I believe (though may be
mistaken) that every MonadTrans is also an instance of a hypothetical
FunctorTrans class, only requiring a Functor constraint to be a
Functor itself; however, ApplicativeTrans doesn't work because StateT
among others requires a Monad constraint to be an Applicative.

On Wed, Jun 2, 2021 at 6:41 PM Carter Schonwald
<carter.schonwald at gmail.com> wrote:
>
> That sounds like a nice idea.  Which laws would we require for it?  The usual monad laws require a pure too right? Along with fmap?
>
> Does this necessitate the existence of applicative trans?
>
> On Wed, Jun 2, 2021 at 12:06 PM Zemyla <zemyla at gmail.com> wrote:
>>
>> I feel like instead, MonadTrans should have a function
>>
>> (>>==) :: Monad m => t m a -> (a -> t m b) -> t m b
>>
>> That way, it can prove it's a Monad while still staying Haskell 98.
>>
>> On Wed, Jun 2, 2021, 10:51 Viktor Dukhovni <ietf-dane at dukhovni.org> wrote:
>>>
>>> On Wed, Jun 02, 2021 at 07:27:28AM +0200, Henning Thielemann wrote:
>>>
>>> > So far, 'transformers' is mostly Haskell 98. This is why I prefer it
>>> > to 'mtl'. Wouldn't it be enough to add this extension to 'mtl'? I see
>>> > that 'mtl' re-uses the MonadTrans class from 'transformers' but maybe
>>> > it should define its own class with the quantified constraints then.
>>>
>>> I don't think that having two incompatible MonadTrans classes would
>>> constitute progress.  Older versions of the transformers library (which
>>> is by now quite stable) will continue to be available, for anyone who
>>> wants to use a Haskell '98 (ish?) version.
>>>
>>> [ FWIW, I don't know what you mean by "is mostly Haskell '98", I'd
>>>   expect that to be a strict binary choice: is or isn't. ]
>>>
>>> --
>>>     Viktor.
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