Documentation

Monads provide substitution (fmap) and renormalization (join):

m >>= f = join (fmap f m)

A free Monad is one that does no work during the normalization step beyond simply grafting the two monadic values together.

[] is not a free Monad (in this sense) because join [[a]] smashes the lists flat.

On the other hand, consider:

data Tree a = Bin (Tree a) (Tree a) | Tip a

instance Monad Tree where
  return = Tip
  Tip a >>= f = f a
  Bin l r >>= f = Bin (l >>= f) (r >>= f)

This Monad is the free Monad of Pair:

data Pair a = Pair a a

And we could make an instance of MonadFree for it directly:

instance MonadFree Pair Tree where
   wrap (Pair l r) = Bin l r

Or we could choose to program with Free Pair instead of Tree and thereby avoid having to define our own Monad instance.

Moreover, Control.Monad.Free.Church provides a MonadFree instance that can improve the asymptotic complexity of code that constructs free monads by effectively reassociating the use of (>>=). You may also want to take a look at the kan-extensions package (http://hackage.haskell.org/package/kan-extensions).

See Free for a more formal definition of the free Monad for a Functor.

A version of lift that can be used with just a Functor for f.

A version of wrap for monad transformers over a free monad.

Note: that this is the default implementation for wrap for MonadFree f (t m).