> If one has f a = Bool -> a, and a value
>
> xs :: f Int
> xs True = 2
> xs False = 3
>
> what are x1 and x2?

One answer I could give is: That depends on the definition of the Traversable instance for the (Bool -> a) type constructor.

Another answer I could give is: Ask the authors of the Traversable documentation. The first sentence on https://hackage.haskell.org/package/base-4.8.1.0/docs/Data-Traversable.html is: "Class of data structures that can be traversed from left to right, ...". So it seems that the authors of that documentation (which I criticize) assume that they can say -- for any given type constructor, so also for your (Bool -> a) example -- what "left" means and what "right" means.


2015-10-24 20:23 GMT+02:00 Matteo Acerbi <matteo.acerbi@gmail.com>:
On Sat, Oct 24, 2015 at 8:12 PM, Janis Voigtländer <janis.voigtlaender@gmail.com> wrote:
It has already been established in this thread what Charles meant by 3.

He meant that a fold-function that has the property he is after would guarantee that it:

a) takes all the content elements from a data structure, say x1,...,xn,

b) builds an application tree with the to-be-folded, binary operation f in the internal nodes of a binary tree, whose leafs, read from left to right, form exactly the sequence x1,...,xn,

c) evaluates that application tree.

Do you agree that what I describe above is a property of a given fold-like function, not of the f handed to that fold-like function?

I might lack some basic knowledge, so thanks for asking.

What does it mean to take all the content elements from a data structure?

If one has f a = Bool -> a, and a value

xs :: f Int
xs True  = 2
xs False = 3

what are x1 and x2?

Best,
Matteo

PS. I won't be able to read the answer before tomorrow. :-)