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What is the Comonad typeclass in Haskell?

What is the Comonad typeclass in Haskell? As in Comonad from Control.Comonad in the comonad package (explanations of any other packages that provide a Comonad typeclass are also welcome). I've vaguely heard about Comonad, but all I really know about it is that is provides extract :: w a -> a, sort of a parallel to Monad's return :: a -> m a.

Bonus points for noting "real life" uses of Comonad in "real" code.

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Dan Burton Avatar asked Dec 08 '11 09:12

Dan Burton


2 Answers

These links may be helpful:

  1. Evaluating cellular automata is comonadic. In particular, "whenever you see large datastructures pieced together from lots of small but similar computations there's a good chance that we're dealing with a comonad".
  2. Sequences, streams, and segments
  3. Comonads in everyday life
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Alexey Romanov Avatar answered Nov 16 '22 10:11

Alexey Romanov


This doesn't fully answer my question, but I wanted to put some relevant information in answer format:

"co" (loosely) means "flip the arrows". Here's a rough visual of that.

Consider the monadic operations:

return :: a ~> m a
flip (>>=) :: (a ~> m b) -> (m a ~> m b)

Reverse the squiggly arrows and you get the comonadic operations:

extract :: a <~ w a
extend :: (a <~ w b) -> (w a <~ w b)

(Written with normal arrows)

extract :: w a -> a
extend :: (w a -> b) -> w a -> w b

Notice how in this format, return is an arrow that just so happens to fit in the argument slot for flip (>>=), and the same is true of extract and extend. Monad/comonad laws say that when you put return or extract into that slot, the result is the identity arrow. The laws are the same, "just with the arrows flipped". That's a super handwavey answer but hopefully it provides some insight.

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Dan Burton Avatar answered Nov 16 '22 10:11

Dan Burton