Comments on: Monads for Free

By: Edward Kmett

Edward Kmett — Wed, 22 Sep 2010 20:27:36 +0000

There are plenty of non-monadic ways to deal with control flow. Applicatives, comonads, continuation passing style, lazy request/response lists, etc.

The Haskell community has converged on monads because out of the alternatives, monads seem to give you the most bang for the buck (and Wadler wrote a lot of the early papers, so they have geek cred.)

By: antivirus 2011

antivirus 2011 — Wed, 22 Sep 2010 01:53:34 +0000

I’m new to this monad stuff, is there other way to control the sequence in which calculations are performed without using monad? Or the compiler/interpreter will strictly decide for that?

By: The Comonad.Reader » Zipping and Unzipping Functors

The Comonad.Reader » Zipping and Unzipping Functors — Sat, 08 Nov 2008 14:16:09 +0000

[...] On the other hand, we do not get the same result for the Free Monad, because it is built over BiffB Either Identity f, and Either is not a zippable bifunctor. [...]

By: Edward Kmett

Edward Kmett — Sun, 13 Apr 2008 07:06:21 +0000

Woops, thats what I get for typing both of those in from memory rather than pasting from the working code. Yes, and yes. Fixed in the post.

I’ll include the cofree comonad next time. I just wanted to put things out in digestible chunks. ;)

The free monad should use a sum, the cofree comonad uses a product.

By: Omar

Omar — Sat, 12 Apr 2008 17:50:44 +0000

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

newtype Free f a = Free { unfree :: (a, f (Free f a)) }

Aren’t those two different types? Isn’t the first a sum and the second a product? Should g = Free f a satisfy g=(f g) + a or g=(f g) x a?

By: Jaak

Jaak — Sat, 12 Apr 2008 15:29:07 +0000

I’m sorry, but what is the Place constructor in the Functor instance of Free? Looks to me that it sould be “Return (f x)” instead.

- Jaak

By: Pseudonym

Pseudonym — Sat, 12 Apr 2008 13:48:40 +0000

And for your next trick, define the free comonad.

Finally, define the free arrow Free a -> Free b.