Re: Proposal: make liftF a method of MonadFree

What about having the class head be

Monad m, Functor f=> … MonadFree f m ..
?

Is the motivation here to have a more performant liftF?

What are some examples of more efficient implementations for current
instances and what’s the performance delta?

On Fri, Jul 16, 2021 at 6:53 PM David Feuer <[email protected]> wrote:

> Another flavor would be to leave liftF alone and add a method that does
> the same thing with a different name. This would preserve performance
> characteristics for instances like FT, for situations where the current
> implementation is faster.
>
> On Fri, Jul 16, 2021, 4:55 PM David Feuer <[email protected]> wrote:
>
>> We have
>>
>> class Monad m => MonadFree f m | m -> f where
>>   wrap :: f (m a) -> m a
>>
>> liftF :: (Functor f, MonadFree f m) => f a -> m a
>> liftF = wrap . fmap pure
>>
>> I propose we change this to
>>
>> class Monad m => MonadFree f m | m -> f where
>>   wrap :: f (m a) -> m a
>>
>>   liftF :: f a -> m a
>>   default liftF :: Functor f => f a -> m a
>>   liftF = wrap . fmap pure
>>
>> and add a function
>>
>> defaultWrap :: MonadFree f m => f (m a) -> m a
>> defaultWrap = join . liftF
>>
>> This change is not strictly backwards compatible. Some instances might,
>> hypothetically, have to add a Functor constraint. For example, the classic
>> Control.Monad.Free and Control.Monad.Trans.Free would need them. However,
>> those instances already have (currently redundant) Functor constraints, so
>> that doesn't seem like a big deal.
>>
>> An alternative would be to hew more strictly to backwards compatibility
>> by placing a Functor f constraint on liftF. This seems a bit sad for
>> "freer" instances that don't need it. For example, we have
>>
>> newtype FT f m a = FT
>>   { runFT :: forall r. (a -> m r) -> (forall x. (x -> m r) -> f x -> m
>> r) -> m r }
>>
>> for which
>>
>> liftF :: f a -> FT f m a
>> liftF fa = FT $ \pur bndf -> bndf pur fa
>>
>> Pull request at https://github.com/ekmett/free/pull/208
>>
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