Just expand out the function composition:
Dual . Endo . flip f = (\x -> Dual (Endo (flip f x)))
which has the type d -> Dual (Endo c).
-- ryan
On 8/26/07, Levi Stephen <levi.stephen@optusnet.com.au> wrote:
Hi,
I was browsing through the source code for Data.Foldable and having trouble
comprehending it (which was kind of the point of browsing the code, so I could
learn something ;) )
I'm looking at foldl
foldl :: (c -> d -> c) -> c -> t d -> c
foldl f z t = appEndo (getDual (foldMap (Dual . Endo . flip f) t)) z
But, I haven't got very far. I'm still trying to follow:
Endo . flip f
f is of type c->d->c, so this makes flip f of type d->c->c.
I think the Endo constructor is of type (a->a)->Endo a
I think a is binding to a function type here, but can not work out what.
(From memory) ghci reports
> :t Endo . flip (+)
Num a => a -> Endo a
So, this looks like a partial application of the constructor?
But, this still isn't helping me understand.
Any thoughts or pointers that might help me comprehend what's happening?
Thanks,
Levi
lstephen.wordpress.com
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