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`

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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