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