Re: [Haskell-cafe] Foldable for (,)

With fmap you can only change all values in some "container". With Foldable you can "fold" it, i.e. calculate some "scalar" result. With Traversable you can "change order of two containers":

sequenceA [[1,2,3],[4,5]] [[1,4],[1,5],[2,4],[2,5],[3,4],[3,5]] sequenceA ("test",[2,3,4]) [("test",2),("test",3),("test",4)] sequenceA ("test",([1,2,3],[4,5,6])) ([1,2,3],("test",[4,5,6]))

2017-05-03 12:12 GMT+03:00 Jonathon Delgado :

Why do you want to traverse a tuple instead of fmap? i.e. what can you do with Foldable/Traversable for (,) that you can't do with Functor?

My background, as you can probably guess, is beginner.

From: Haskell-Cafe on behalf of Chris Smith Sent: 03 May 2017 08:51 To: Tony Morris Cc: haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] Foldable for (,)

Replying to myself, I suppose one good answer is that whether or not you care about Foldable instances for tuples, you might care about Traversable instances, and those require Foldable as a superclass.

For example, one possible specialization of `traverse` is:

traverse :: (a -> IO b) -> (SideValue, a) -> IO (SideValue, b)

Jonathon, I don't know how much background you're coming from, so I'd be happy to explain that in more detail if you need it.

On Wed, May 3, 2017 at 1:44 AM, Chris Smith wrote:

I'm also interested in Jonathon's question, so let me try to bring things back to the question. Everyone agrees that there's only one reasonable way to define this instance if it exists. But the question is: why is it defined at all?

That's an easy question to answer for Functor, Applicative, and Monad. But I am having trouble giving a simple or accessible answer for Foldable. Do you know one?

On Wed, May 3, 2017 at 1:32 AM, Tony Morris wrote: It's Foldable for ((,) a).

It is not Foldable for any of these things:

* (,) * tuples * pairs

In fact, to talk about a Foldable for (,) or "tuples" is itself a kind error. There is no good English name for the type constructor ((,) a) which I suspect, along with being unfamiliar with utilising the practical purpose of types (and types of types) is the root cause of all the confusion in this discussion.

Ask yourself what the length of this value is:

[[1,2,3], [4,5,6]]

Is it 6? What about this one:

[(1, 'a'), (undefined, 77)]

Is it 4? No, obviously not, which we can determine by:

:kind Foldable :: (* -> *) -> Constraint :kind [] :: * -> *

Therefore, there is no possible way that the Foldable instance for [] can inspect the elements (and determine that they are pairs in this case). By this method, we conclude that the length of the value is 2. It cannot be anything else, some assumptions about length itself put aside.

By this ubiquitous and very practical method of reasoning, the length of any ((,) a) is not only one, but very obviously so.

On 03/05/17 17:21, Jonathon Delgado wrote:

...

I sent the following post to the Beginners list a couple of weeks ago (which failed to furnish an actual concrete example that answered the question). Upon request I'm reposting it to Café:

I've seen many threads, including the one going on now, about why we need to have:

length (2,3) = 1 product (2,3) = 3 sum (2,3) = 3 or (True,False) = False

but the justifications all go over my head. Is there a beginner-friendly explanation for why such seemingly unintuitive operations should be allowed by default? _______________________________________________ Haskell-Cafe mailing list To (un)subscribe, modify options or view archives go to: http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe Only members subscribed via the mailman list are allowed to post.

_______________________________________________ Haskell-Cafe mailing list To (un)subscribe, modify options or view archives go to: http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe Only members subscribed via the mailman list are allowed to post.

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