[Haskell-beginners] Foldable of Foldable
julian.birch at gmail.com
Fri Jan 2 16:28:09 UTC 2015
Thankyou Tony, that works perfectly. I think my biggest problem was that I
didn't even know you could write
newtype (Compose g f) a = O (g (f a))
It's logical, but I couldn't figure it out and none of the web pages I
could find mentioned it (and the compiler just complains if you try
"newtype Compose g f a = "
Also, is the Functor declaration really necessary? Seems like you can
instance (Foldable f1, Foldable f2) => Foldable (Compose f1 f2) where
foldr f start (O list) = foldr g start list
where g = flip . foldr f
I'm certainly going to spend some time examining Control.Compose. I think
my Haskell brain bending just went up a level, (sadly, not my actual brain.)
On 2 January 2015 at 15:16, Tony Morris <tonymorris at gmail.com> wrote:
> A foldable of a foldable is itself foldable. That fact is expressed in
> instance (Foldable g, Foldable f, Functor g) => Foldable (:. g f)
> On 03/01/15 01:14, Julian Birch wrote:
> Apologies if this isn't clear, I suspect if I understood the terminology
> better I'd already have solved the problem. I've got a foldable of a
> foldable of a. (Specifically `[Set a]`) and I want to be able to express
> the concept that I can treat `(Foldable f1, Foldable f2) => f1 (f2 a)` as
> `f3 a` i.e. a foldable of a foldable of a can be newtyped to a foldable of
> a. At least, I think that's right.
> Sadly, my attempts at actually expressing this concept have met with
> incomprehension from GHC, could someone help me out, please?
> Beginners mailing listBeginners at haskell.orghttp://www.haskell.org/mailman/listinfo/beginners
> Beginners mailing list
> Beginners at haskell.org
-------------- next part --------------
An HTML attachment was scrubbed...
More information about the Beginners