What's confusing here is how currying works with infix operators. It's
idiomatic in Elm to have your accumulator be the last argument, and, for
instance, if you were writing your own data type, you would want to write
your functions so that they can be chained together easily:
myMatrix
|> scale 2
|> subtract 5
|> subtractMatrix myOtherMatrix
|> normalize
But as an infix operator (-) is not able to follow that convention;
5
|> (-) 3
|> (-) 1
is confusingly equivalent to `(1 - (3 - 5))` rather than to `5 - 3 - 1`
If you had a function `subtract` such that
5 |> subtract 3 |> subtract 1 == (5 - 3 - 1)
then you could use that function with fold as you intend
List.foldl subtract 0 [1, 2, 3, 4] == -10
You can achieve the same result with
List.foldl (flip (-)) 0 [1, 2, 3, 4] == -10
Another way to put it is, in Elm, folds expand in the following way:
List.foldl f x [b, c, d] == x |> f b |> f c |> f d
List.foldr f x [b, c, d] == f b <| f c <| f d <| x
On Thu, Dec 8, 2016 at 7:50 PM, Kasey Speakman <kjspeak...@gmail.com> wrote:
> (deleted and corrected original post with proper expansion of Elm's foldl)
>
> I know this is a really old thread, but I ran into this precise question
> and thought I would add a perspective.
>
> The form a -> b -> b is not left-building, regardless of the direction you
> are traversing the list.
>
> An example: Starting from zero, subtract the numbers 1, 2, and 3. The
> expected answer is -6.
>
> List.foldl (-) 0 [1, 2, 3]
> -> returns -6 in Haskell (well, actually tested in F# which uses same
> order as Haskell)
> expands to: ((0 - 1) - 2) - 3 = -6
> -> returns 2 in Elm
> expands to: 3 - ((1 - 0) - 2)
>
> Elm's expansion is wonky for this. It appears to be center-building:
> List.foldl (-) 0 [1] -- returns 1, expands 1 - 0
> List.foldl (-) 0 [1, 2] -- returns -1, expands (1 - 0) - 2
> List.foldl (-) 0 [1, 2, 3] -- returns 2, expands 3 - ((1 - 0) - 2)
> List.foldl (-) 0 [1, 2, 3, 4] -- returns -2, expands (3 - ((1 - 0) -
> 2)) - 4
>
> When a and b are the same type it will only return the correct answer if
> the fold operation is also commutative or if flip is used to correct the
> ordering. When a and b are not the same type, the compiler will provide an
> error for wrong ordering of course.
>
> I started out on the side that a -> b -> b was correct as that feels like
> proper "reduction" or chainable syntax. But after exploring it, it is
> clearly not left-building. Makes sense when you consider this form is used
> with pipe to convert right-building operations into left-reading code. e.g. a
> |> f |> g |> h instead of h (g (f a))
>
> On Tuesday, July 16, 2013 at 6:13:01 AM UTC-5, Evan wrote:
>>
>> Gotcha, I definitely see the reasoning :)
>>
>>
>> On Tue, Jul 16, 2013 at 12:54 PM, Balazs Komuves <bkom...@gmail.com>
>> wrote:
>>
>>>
>>> I was not engaging in debate, religious or not (though I tend to have
>>> very strong opinions about these questions). I was explaining why I think
>>> Haskell uses the order it uses (because it is distinguished from a
>>> mathematical viewpoint). Of course you are not required to follow that
>>> convention, I was just pointing out that it is not simply an ad-hoc choice.
>>>
>>> Balazs
>>>
>>>
>>>
>>> On Tue, Jul 16, 2013 at 12:21 PM, Evan Czaplicki <eva...@gmail.com>
>>> wrote:
>>>
>>>> I think this might be a religious debate on some level. My first
>>>> functional languages were Scheme
>>>> <http://docs.racket-lang.org/reference/pairs.html#(def._((lib._racket/private/list..rkt)._foldl))>
>>>> and Standard ML <http://www.standardml.org/Basis/list.html>. The
>>>> libraries I just linked both use the same argument order for foldl and
>>>> foldr as in Elm. I was raised a certain way and it just stuck in my mind. I
>>>> suspect that everyone prefers the order they learned first because it
>>>> matches their mental model.
>>>>
>>>> I wrote up a bunch of "reasoning", but really, I am just engaging in
>>>> the religious debate. I'd feel bad deleting it all though, so here is some
>>>> of it:
>>>>
>>>> OCaml's list library
>>>> <http://caml.inria.fr/pub/docs/manual-ocaml/libref/List.html> does it
>>>> the way you suggest. I find this order offensive on some level.
>>>>
>>>> The big questions for "physical" argument order are as follows:
>>>>
>>>> - What is the type of `fold` or `reduce`? When you fold an
>>>> unordered thing, is it from the right or the left?
>>>> - What is the type of `foldp`? Which way does time go? Is this
>>>> cultural?
>>>>
>>>> I don't find these questions particularly useful, and I don't think
>>>> programmers should have to wonder about them to use fold and foldp.
>>>>
>>>> At the end of the day, I chose the types on purpose. I find them easier
>>>> to use, easier to teach, easier to understand. I want to keep them this
>>>> way.
>>>>
>>>>
>>>> On Tue, Jul 16, 2013 at 10:40 AM, Balazs Komuves <bkom...@gmail.com>
>>>> wrote:
>>>>
>>>>>
>>>>> The Haskell version of the foldl is the "right one" in the following
>>>>> sense:
>>>>>
>>>>> foldl makes sense in general for left-associative operators, and foldr
>>>>> makes sense for right-associative operators.
>>>>> Left-associative operators must have the type (a -> b -> a), while
>>>>> right-associative operators must have type (a -> b -> b).
>>>>>
>>>>> I think the fact that you cannot change a foldr to foldl without
>>>>> changing the types is actually an advantage: it forces you to think about
>>>>> which version is the "proper" one, and you cannot accidentally do the
>>>>> wrong
>>>>> one. Of course sometimes it can be inconvenient.
>>>>>
>>>>> What I somewhat dislike in the Haskell version of foldr (not foldl),
>>>>> is that while
>>>>>
>>>>> (foldl . foldl . foldl) etc makes sense, (foldr . foldr) does not; for
>>>>> that to work you would have to flip the last two arguments:
>>>>>
>>>>> myfoldr :: (a -> b -> b) -> ([a] -> b -> b)
>>>>> myfoldr f xs y = foldr f y xs
>>>>>
>>>>> But the practicality of this change is debatable, I guess.
>>>>>
>>>>> Balazs
>>>>>
>>>>>
>>>>>
>>>>>
>>>>> On Wed, Jul 10, 2013 at 4:38 PM, Evan Czaplicki <eva...@gmail.com>
>>>>> wrote:
>>>>>
>>>>>> It's partly about composability (i.e. the data structure should be
>>>>>> last).
>>>>>>
>>>>>> It is also about reuse. In Elm it is valid to say:
>>>>>>
>>>>>> foldl (::) []
>>>>>> foldr (::) []
>>>>>>
>>>>>> If I want to change the order of my traversal, I should not *also* need
>>>>>> to change the definition of mildly related functions or start using
>>>>>> flip on things.
>>>>>>
>>>>>> Finally, once you know that the accumulator is always the second
>>>>>> argument, you do not have to look at docs anymore. Even now I forget the
>>>>>> order of arguments in Haskell's folds and need to look it up.
>>>>>>
>>>>>> I first learned this way from Standard ML
>>>>>> <http://www.standardml.org/Basis/list.html>, and it is my favorite
>>>>>> by far.
>>>>>>
>>>>>>
>>>>>> On Wed, Jul 10, 2013 at 4:12 PM, Tim hobbs <tim.t...@gmail.com>
>>>>>> wrote:
>>>>>>
>>>>>>> Well, elm's ordering is more useful. For example, I recently had a
>>>>>>> case where I wrote:
>>>>>>>
>>>>>>> let
>>>>>>> irrelivantFuncitonName fold = fold blabla default list
>>>>>>> in
>>>>>>> irrelivantFunctionName foldl + irrelivantFuncitonName foldr
>>>>>>>
>>>>>>> In Haskell, the same example ends up being
>>>>>>>
>>>>>>> let
>>>>>>> irrelivantFuncitonName fold = fold blabla default list
>>>>>>> in
>>>>>>> irrelivantFunctionName foldl + irrelivantFuncitonName (\f d l->
>>>>>>> foldr (\a b->f b a) d l)
>>>>>>>
>>>>>>> Tim
>>>>>>>
>>>>>>>
>>>>>>> On Wednesday, July 10, 2013 4:03:23 PM UTC+2, Zsombor Nagy wrote:
>>>>>>>>
>>>>>>>> Hi!
>>>>>>>>
>>>>>>>> I wonder why is the foldl in Elm and in Haskell calling the binary
>>>>>>>> operator with arguments in a different order?
>>>>>>>>
>>>>>>>> foldl (\t acc -> acc + 1) 0 [1, 1, 1, 1, 1, 1]
>>>>>>>> haskell: 2
>>>>>>>> Elm: 6
>>>>>>>>
>>>>>>>> For me the haskell way seems more straightforward, but maybe that
>>>>>>>> "optimal composibility guideline" makes this turn around?
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>> zs
>>>>>>>
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