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

Heinrich Apfelmus apfelmus at quantentunnel.de
Sun Dec 1 16:06:10 UTC 2013

Nickolay Kudasov wrote:
>> Well, that they are heavily used in practice does not mean that they are
>> actually useful in practice. The thing is that as soon as your functor is a
>> monad, I don't think you really need the Free type anymore -- your
>> instruction type is already a monad.
> I don't​​ use that myself, so I leave this for others to answer. But you
> should note that `Free m` is not the same as `m`: e.g. if `m` is a
> probability monad `newtype P a = P [(Double, a)]`, then `Free P` gives you
> much more: the whole tree of probabilities (not only probs of final
> results), so one could traverse that tree. So I believe `Free m` is rather
> useful (as is deriving instances for `Free m` the way it is).

Yes, but in this case,  Free P  is useful *regardless* of  P  being a 
monad. If you didn't declare a monad instance for the  P  type, then you 
would still get the tree of probabilities.

It bugs me that the functor is expected to already be a monad. The instance

     MonadState s m => MonadState s (Free m)

is a fancy way of saying that if the instruction type  m  has two 
instructions  get  and  put , then the free monad will have them as 
well. This is fine from the perspective of variant types, but we don't 
need to know that  m  is a monad for that. Furthermore, the  MonadState 
  class suggests that the usual laws for the state monad hold -- which 
they do not! Here a counterexample:

     {-# LANGUAGE FlexibleInstances, MultiParamTypeClasses #-}
     import Control.Monad.State

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

     instance Functor f => Monad (Free f) where
         return         = Return
         (Free f) >>= k = Free $ fmap (>>= k) f

     instance MonadState s (Free (State s)) where
         state f = Free . state $ \s ->
             let (a,s') = f s in (Return a, s')

     interpret :: Free (State s) a -> (s -> s)
     interpret (Return a) s0 = s0
     interpret (Free f  ) s0 = snd (runState f s0)
         -- apply only the first instruction, skip the rest

     example1 = interpret (put 'a' >> get >> put 'b' >> get) undefined
     example2 = interpret (put 'b' >> get) undefined

If we expect the usual laws for the state monad, then both  example1 
and  example2  should be the same value. However, this is not the case: 
  example1 = 'a'  while  example2 = 'b' .

Just because you have two operations  put  and  get  doesn't mean that 
they can't have additional effects. And apparently, the  MonadState 
condition is not strong enough to guarantee all the put/get laws.

>> (But I think it will be impossible for MonadCont).
> It is. See https://github.com/ekmett/free/pull/33 for FreeT. FT has the
> instance in HEAD already.

That's surprising, I will have to check that.

It appears to me that the  MonadReader  instance is only correct because 
the control operation is a morphism:

    local f (m >>= k)  =  local f m >>= local f . k

Otherwise, I don't see how a general control operation can be lifted.

>> Almost, but not quite. The key qualification is "while still allowing
>> pattern matching". 
> You're​​ right. But I think it is unnecessary for a library user to pattern
> match on F's structure. It is pattern matching on supplied functor that
> matters. And that ability is not lost.

A pattern match

    view :: F f a -> Either a (f (F f a))

that runs in O(1) time is very useful for implementing interpreters. For 
an example, see [1]. In particular, we can use the remainder to create 
new monadic actions with (>>=).

The distinction is similar between being able to pattern match on (:) 
and using only  fold  to operate on lists. The former is more flexible.


> To summarize, I currently don't see what 'free' offers that the
>> 'operational' package can't do equally well with only 11 exported symbols..
> As far as I can tell, while with operational you can certainly do more
> things, free provides more things for free (these "baked algebraic laws").
> free also provides some other interesting things, like iterative (co)monad
> trasformers, cofree comonads and free applicatives/alternatives (which are
> out of operational/free common area).
> That all said, I don't feel myself concerned/experienced enough to state
> that one package should be preferred to another.

As mentioned, I'm not a fan of the "baked in algrebraic laws" because 
this excludes an optimization and many laws have to be written in the 
interpreter anyway.

But you're saying that 'free' provides other free structures besides 
monads. That's a good point.

Best regards,
Heinrich Apfelmus


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