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
write
```
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.)
Thanks again,
Julian.
On 2 January 2015 at 15:16, Tony Morris
A foldable of a foldable is itself foldable. That fact is expressed in Control.Compose:
https://hackage.haskell.org/package/TypeCompose-0.9.10/docs/Control-Compose....
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?
Thanks,
Julian.
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