Out of curiosity, why do you require the `MonadIO` on the `Monad`
instance?
Thanks Tom and Rodrigo.
That clarifies the problem. We will need to think which solution makes better sense.
On Wed, 12 Apr 2023 at 15:01, Rodrigo Mesquita <rodrigo.m.mesquita@gmail.com> wrote:
Indeed, this is included in the GHC 9.6.x Migration Guide._______________________________________________
Unfortunately, I’m also not sure there is a solution for this particular where (T m) is only a Monad if m instances MonadIO.As Tom explained, under transformers 0.6 `T` no longer is a monad transformer.
A few workarounds I can think of:
- No longer instance `MonadTrans T`, and use a instance `MonadIO m => MonadIO (T m)` instead.Rationale: if you always require `m` to be `MonadIO`, perhaps the ability to always lift an `m` to `T m` with `liftIO` is sufficient.
- Add the `MonadIO` instance to the `m` field of `T`, GADT style, `data T m a where T :: MonadIO m => m -> T m a`Rational: You would no longer need `MonadIO` in the `Monad` instance, which will make it possible to instance `MonadTrans`.
- Redefine your own `lift` regardless of `MonadTrans`
Good luck!Rodrigo
On 12 Apr 2023, at 10:10, Tom Ellis <tom-lists-haskell-cafe-2017@jaguarpaw.co.uk> wrote:
On Wed, Apr 12, 2023 at 02:32:43PM +0530, Harendra Kumar wrote:
instance MonadIO m => Monad (T m) where
return = pure
(>>=) = undefined
instance MonadTrans T where
lift = undefined
I guess it's nothing to do with 9.6 per se, but rather the difference
between
* https://hackage.haskell.org/package/transformers-0.5.6.2/docs/Control-Monad-Trans-Class.html#t:MonadTrans
* https://hackage.haskell.org/package/transformers-0.6.1.0/docs/Control-Monad-Trans-Class.html#t:MonadTrans
I'm not sure I can see any solution for this. A monad transformer `T`
must give rise to a monad `T m` regardless of what `m` is. If `T m`
is only a monad when `MonadIO m` then `T` can't be a monad transformer
(under transformers 0.6).
Tom
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