What was wrong with Control.MonadPlus.Free?

Question

The free MonadPlus defined as

data Free f a = Pure a | Free (f (Free f a)) | Plus [Free f a]

was removed in free 4.6 with the foll

Answer 1

I believe that the representation itself was OK and that the lawfulness could have been remedied by changing these method signatures

iter :: Functor f => (f a -> a) -> ([a] -> a) -> Free f a -> a
iterM :: (Monad m, Functor f) => (f (m a) -> m a) -> ([m a] -> m a) -> Free f a -> m a

to

iter :: (Functor f, Monoid a) => (f a -> a) -> Free f a -> a
iterM :: (MonadPlus m, Functor f) => (f (m a) -> m a) -> Free f a -> m a

i.e.

• use Monoid a instead of an arbitrary function [a] -> a in iter;
• use MonadPlus m instead of an arbitrary function [m a] -> m a in iterM.

My guess is that it was removed (instead of fixed) just because it is not worth to keep around when FreeT f [] gives an equivalent representation.