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

Foldable is required tor datatype to be a Traversable: class (Functor t, Foldable t) => Traversable (t :: * -> *) ((,) a) is Functor. To be Traversable it has to be Foldable. If ((,) a) is Traversable then we can write: sequence (1, Just 2) -- == Just (1,2) sequence (1, Nothing) -- == Nothing There could be other useful classes or functions which required Foldable. An other side, a tuple (with parameterized second part) can be a part of complex datatype and possibly we need Foldable or Traversable instance for that type. If someone inhabits to think about tuple as a Functor, he/she can think about tuple as Foldable and Traversable as well: fmap (+1) (1,2) == (1,3) foldMap (+1) (1,2) == 3 There are other datatypes with similar Foldable instances. I mean a least Identity. length (Identity [1,2,3]) == 1 2017-05-03 11:41 GMT+03:00 Jonathon Delgado :

Thank you for your explanation, but I think I'm missing something basic. Lists can have a variable length, so it makes sense to have operations that return the length or operate over a set. As ((,) a) can only have one value, the Foldable operations appear to be redundant as well as misleading (by implying that there could be more than one value).

From: Haskell-Cafe on behalf of Tony Morris Sent: 03 May 2017 08:32 To: haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] Foldable for (,)

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:

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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

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