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
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. :-)