问题
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.
回答1:
These links may be helpful:
- 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".
- Sequences, streams, and segments
- Comonads in everyday life
回答2:
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.
来源：https://stackoverflow.com/questions/8428554/what-is-the-comonad-typeclass-in-haskell