# [Haskell-cafe] free vs. operational vs. free-operational

Heinrich Apfelmus apfelmus at quantentunnel.de
Sat Nov 30 11:03:46 UTC 2013

```Alejandro Serrano Mena wrote:
> Dear Café,
> I've been reading lately about the relation between the 'free' package and
> the 'operational' package for rolling your own monads [1] [2]. Furthermore,
> I've discovered that the 'free-operational' package, which is some sort of
> bridge between the two worlds, and provides not only Monad but also
> Applicative and Alternative instances for ProgramT.
> The problem is that right now everything is a little confused in my head.
> In particular, I have the following questions:

(Author of 'operational' here.)

> - I've read that free allows you to 'bake algebraic laws' in the resulting
> monad. How does this work? Why doesn't operational offer that feature?

What I mean by 'baking in algebraic laws' is the following: Consider the

data F a = MZero | MPlus a a

mzero :: Free F a
mzero = Free MZero

mplus :: Free F a -> Free F a -> Free F a
mplus x y = Free (MPlus x y)

For convenience, let me reproduce the relevant definitions for the free

data Free f a = Return a | Free (f (Free f a))

(>>=) :: Functor f => Free f a -> (a -> Free f b) -> Free f b
(>>=) (Return a) k = k a
(>>=) (Free   x) k = Free (fmap (>>= k) x)

Now, if you think about the definition of bind for a moment, you will
see that it automatically guarantees a distributive law for  mplus :

mplus x y >>= k  =  mplus (x >>= k) (y >>= k)

However, it turns out [1] that there is another law that you might want
mplus  to satisfy

mplus (return a) y = return a

but which is incompatible with the distributive law. So, if you want to
implement a monad where  mplus  should obey the latter law, you have to

In the 'free' approach, I find it unpleasant that some laws are
automatic from the functor type, while others have to be ensured by the
interpreter. That's why 'operational' offers only one way to implement
monads: everything has to be done in the interpreter.

> - One of the things I really like from the free package is that it provides
> support for many monad transformer stacks, whereas operational does not? Is
> there any special restriction why operational cannot do this? Would it be

There is a good reason why 'operational' cannot do this: in general, it
is impossible to mix different effects in a general way. Why would

ProgramT SomeInstruction (State s)

be a state monad as well even though  SomeInstruction  can introduce new
effects?

If you look at the monad transformer instances for  Free , like
MonadState, you will notice that they require the functor to be that
monad, i.e. they make use of the "baking in laws" effect. This is quite
useless in practice, as writing a MonadState instance of the instruction
type F is the same work as writing a MonadState instance for the  Free F

If you look at the transformer version  Control.Monad.Trans.Free , you
will see that there are no MonadState instances -- as expected, because
you have to specify the interaction of effects.

> - It seems that free gives more features (Alternative, Applicative) with
> the same work. In which situations should I prefer operational to free? I
> really like the separation between providing a data type and then a
> interpretation that operational embodies...

Well, the features may look good on screen, but once you check the
preconditions for the class instances, you will find that fulfilling
them is as much work as writing the instance from scratch.

The only two things that a free monad can give you is:

* A way to pattern match on instructions and write an interpreter.

This is what operational does. Everything else just shuffles work
around, but doesn't alleviate it for you.

That said, as we saw, Free can give you some laws automatically.
However, this also has a drawback: Program has an optimization that Free
can never have. Namely, Program gives you a (>>=) that can be used in a
left-associative way (Think (((x ++ y) ++ z) ++ w) ) while still
allowing pattern matching.

> - Should I replace my usage of operational with free-operational altogether?

I would say no, but then again, I'm the author of the 'operational'
package. :)

Best regards,
Heinrich Apfelmus

--
http://apfelmus.nfshost.com

```