[Haskell-cafe] 'Associative' order of calling

Matteo Acerbi matteo.acerbi at gmail.com
Sun Oct 25 09:09:21 UTC 2015

Janis Voigtländer writes:

> 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. [..]

Since Charles had not mentioned Traversable in his messages, I still
find the original question 2 to be ambiguous.

If one has a Traversable instance, then he can say what being left
fold-like means for a function. As far as I understand, however, this
should be equivalent to the property of being extensionally equal to the
foldl1 of the Foldable superclass, provided instances follow the laws.

Please, again, correct me if I am wrong.

I want to remark that mine is not criticism towards Charles's message:
on the contrary, I think that starting a discussion on not completely
specified ideas can be very beneficial for everyone, as several people
at once will make some effort to get a clearer understanding, sharing
their points of view.

Thanks for taking the time to explain your interpretation.


More information about the Haskell-Cafe mailing list