Le 23 oct. 2012 09:54, "Alfredo Di Napoli" <alfredo.dinapoli@gmail.com> a écrit :
>
> What this code does is straighforward. I was struck from the following sentences in LYAH: 
>
>> Notice that we didn't have to provide the function that takes a value and returns a monoid value. 
>> We receive that function as a parameter to foldMap and all we have to decide is where to apply 
>> that function and how to join up the resulting monoids from it.
>
>
> This is obvious, so in case of (+) f = Sum, so f 3 = Sum 3 which is a Monoid.
> What I was wondering about is how Haskell knows that it has to pass, for example, Sum in case of (+) and Product in case of (*)?
> In other term, I'm missing the logical piece of the puzzle which maps the associative binary function passed to fold up to our foldMap declaration.

Haskell doesn't know it has to pass anything ! As the doc said, this is a parameter to foldmap : so you would use "treeSum = foldmap Sum" for example. And the binary function associed would be determined by the monoid instance. Here that would be (+) because that's what mappend is for Sum (or rather for Sum Int).

--
Jedaï