Enum.reduce + :lists.reverse() vs List.foldr

[mraptor]

hi,

Why would I use :
Enum.reduce + :lists.reverse()
instead of :
List.foldr

[Rob Lally]

Because Enum.reduce will work on any enumerable - such as a stream, and List.foldr works on Lists.

R,

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[Onorio Catenacci]

It's also worth considering that some functions may not yield the same reduction depending on the order in which the elements are seen.  Consider this contrived example:

iex(1)> [1,2,3,4] |> List.foldr(0,&(&1 - &2))

-2

vs.

iex(2)> [1,2,3,4] |> List.foldl(0,&(&1 - &2))

2

In this particular case, Enum.reduce will act like a foldl. And I'm sure there are plenty of other operations where order of traversal of the list is going to be important as well.

FWIW.

--

Onorio

[Onorio Catenacci]

Wow, I'm a bit embarrassed that I wasn't a bit more careful about replying there.

While I am sure the point about the order in which elements of a list are seen is a valid one, it looks like I've somehow set up that example incorrectly.  In this particular case, either way fold should yield -10. I'm not quite sure how I managed to flummox that up so badly.

So, anyone--please help me wipe the egg from my face--what did I do wrong in typing in those expressions?  Why did I get two different results and why did neither of them come out to -10 (seemingly the correct answer)?

--

Onorio

[Peter Hamilton]

Your function is called with (el, acc), not (acc, el).

1 - 0 = 1
2 - 1 = 1
3 - 1 = 2
4 - 2 = 2

And

4 - 0 = 4
3 - 4 = -1
2 - -1 = 3
1 - 3 = -2

[Onorio Catenacci]

Thanks Peter.  That makes perfectly good sense now.