Am I missing something about unfoldr?

[David Feuer]

Hmm... I found something I missed. I need to prove that this doesn't break
when the arguments to unfold use seq (if in fact it doesn't). I will have
to look into it later.
On Jul 26, 2014 3:34 PM, "David Feuer" <david.feuer at gmail.com> wrote:

> The current definition:
>
> unfoldr :: (b -> Maybe (a, b)) -> b -> [a]
> unfoldr f b = case f b of
>   Just (a,new_b) -> a : unfoldr f new_b
>   Nothing -> []
>
> I haven't yet looked over the Core to see if GHC is able to do a
> worker/wrapper transformation to inline it, but we definitely want to
> inline it (in some appropriate phase), because doing so has the potential
> to erase the Maybes. It also isn't a good producer for foldr/build.
> However, if we (re)write it like this:
>
> unfoldrB f b c n = go b
>   where
>     go b = case f b of
>              Just (a,new_b) -> a `c` go new_b
>              Nothing -> n
>
> unfoldr f b = build (unfoldrB f b)
>
> then we get some really nice fusion:
>
> foldr c n (unfoldr f b)
> = foldr c n (build (unfoldrB f b))
> = unfoldrB f b c n
>
> The list is gone, and there are no new closures or data structures in its
> place.
>
> As a side benefit, we could write groupBy as an unfoldr, and make it a
> somewhat better producer, especially when the groups are small. I don't
> *think* there's any way to make groupBy a good consumer for foldr/build,
> unfortunately.
>
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