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
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 _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe