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:

...

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

Haskell-Cafe Info Page mail.haskell.org This mailing list is for the discussion of topics related to Haskell. The volume may at times be high, as the scope is broader than the main Haskell mailing list.

...

Only members subscribed via the mailman list are allowed to post.

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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.

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.

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.

_______________________________________________ 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.

Look how instance for List is defined.

instance Traversable https://hackage.haskell.org/package/base-4.9.1.0/docs/src/Data.Traversable.h... [] where {-# INLINE traverse #-} -- so that traverse can fuse traverse https://hackage.haskell.org/package/base-4.9.1.0/docs/src/Data.Traversable.h... f https://hackage.haskell.org/package/base-4.9.1.0/docs/src/Data.Traversable.h... = List.foldr https://hackage.haskell.org/package/base-4.9.1.0/docs/src/GHC.Base.html#fold... cons_f https://hackage.haskell.org/package/base-4.9.1.0/docs/src/Data.Traversable.h... (pure https://hackage.haskell.org/package/base-4.9.1.0/docs/src/GHC.Base.html#pure []) where cons_f https://hackage.haskell.org/package/base-4.9.1.0/docs/src/Data.Traversable.h... x https://hackage.haskell.org/package/base-4.9.1.0/docs/src/Data.Traversable.h... ys https://hackage.haskell.org/package/base-4.9.1.0/docs/src/Data.Traversable.h... = (:) <$> https://hackage.haskell.org/package/base-4.9.1.0/docs/src/Data.Functor.html#... f https://hackage.haskell.org/package/base-4.9.1.0/docs/src/Data.Traversable.h... x https://hackage.haskell.org/package/base-4.9.1.0/docs/src/Data.Traversable.h... <*> https://hackage.haskell.org/package/base-4.9.1.0/docs/src/GHC.Base.html#%3C%... ys https://hackage.haskell.org/package/base-4.9.1.0/docs/src/Data.Traversable.h...

It uses List.foldr. Many other instances do the same.

Functions in all instances of class should have the same signatures. So we have to add Foldable constraint to the class. Of cause we can implement 'foldr' internaly in 'traverse' if needed (as well as fmap). But this is not so good and more important that in this case we don't know how to derive Traversable instances automatically. So the answer - many instances wouldn't compile and DeriveTraversable wouldn't work. 2017-05-03 12:56 GMT+03:00 Jonathon Delgado :

OK, I understand why Traversable is useful here - thank you Chris and Dmitry!

The next question is why Traversable requires Foldable. I looked at the source, and couldn't see where Foldable is being used, other than as a constraint on Traversable. To put the question differently, what would fail to compile if this constraint was removed?

From: Dmitry Olshansky Sent: 03 May 2017 09:53 To: Jonathon Delgado Cc: haskell-cafe@haskell.org Subject: 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.

_______________________________________________ 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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https://i.imgur.com/A2enuhq.png On 03/05/17 21:17, Jonathon Delgado wrote:

List.foldr has signature (a -> b -> b) -> b -> [a] -> b, i.e. an actual list? How is this effected by the Foldable constraint?

From: Dmitry Olshansky Sent: 03 May 2017 10:47 To: Jonathon Delgado Cc: haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] Foldable for (,)

Look how instance for List is defined.

instance Traversable [] where {-# INLINE traverse #-} -- so that traverse can fuse traverse f = List.foldr cons_f (pure []) where cons_f x ys = (:) <$> f x <*> ys It uses List.foldr. Many other instances do the same. Functions in all instances of class should have the same signatures. So we have to add Foldable constraint to the class. Of cause we can implement 'foldr' internaly in 'traverse' if needed (as well as fmap). But this is not so good and more important that in this case we don't know how to derive Traversable instances automatically.

So the answer - many instances wouldn't compile and DeriveTraversable wouldn't work.

2017-05-03 12:56 GMT+03:00 Jonathon Delgado : OK, I understand why Traversable is useful here - thank you Chris and Dmitry!

The next question is why Traversable requires Foldable. I looked at the source, and couldn't see where Foldable is being used, other than as a constraint on Traversable. To put the question differently, what would fail to compile if this constraint was removed?

From: Dmitry Olshansky Sent: 03 May 2017 09:53 To: Jonathon Delgado

Cc: haskell-cafe@haskell.org Subject: 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.

_______________________________________________ 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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Traversable instances (for "singletons" or not) can be implemented _uniformly_ using Functor and Foldable instances. And they are implemented in such way when you derive Traversable. So we need Foldable constraint for the class. E.g. {-# LANGUAGE DeriveFunctor, DeriveFoldable,DeriveTraversable #-}

data D a = D { f1 :: [a], f2 :: (String, a) } deriving (Show, Eq, Functor, Foldable, Traversable) sum $ D [1..3] ("test",4) 10 mapM_ print $ sequenceA $ D [[1,2],[3,4]] ("test",[5,6]) D {f1 = [1,3], f2 = ("test",5)} D {f1 = [1,3], f2 = ("test",6)} D {f1 = [1,4], f2 = ("test",5)} D {f1 = [1,4], f2 = ("test",6)} D {f1 = [2,3], f2 = ("test",5)} D {f1 = [2,3], f2 = ("test",6)} D {f1 = [2,4], f2 = ("test",5)} D {f1 = [2,4], f2 = ("test",6)}

2017-05-03 14:34 GMT+03:00 Jonathon Delgado :

Interesting, that's not the one linked to from Dmitry's code.

In any case, is this correct?

1) Traversable is useful for all containers, including ones which can only hold a single value, such as (,) a. 2) The traversable definition for containers which can hold multiple versions requires Foldable. 3) So Traversable has to depend on Foldable. 4) So singleton containers also have to implement Foldable, even when it doesn't really make sense to do so.

Is there some kind of refactoring which would "fix" this, other than two unrelated Traversable classes? I understand that it might be impractical to refactor a widely-used standard library, but I would be interested in how such a situation could be avoided when starting from scratch.

From: Haskell-Cafe on behalf of Tony Morris Sent: 03 May 2017 11:21 To: haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] Foldable for (,)

https://i.imgur.com/A2enuhq.png

...

List.foldr has signature (a -> b -> b) -> b -> [a] -> b, i.e. an actual

On 03/05/17 21:17, Jonathon Delgado wrote: list? How is this effected by the Foldable constraint?

...

From: Dmitry Olshansky Sent: 03 May 2017 10:47 To: Jonathon Delgado Cc: haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] Foldable for (,)

Look how instance for List is defined.

instance Traversable [] where {-# INLINE traverse #-} -- so that

...

Functions in all instances of class should have the same signatures. So we have to add Foldable constraint to the class. Of cause we can implement 'foldr' internaly in 'traverse' if needed (as well as fmap). But this is not so good and more important that in this case we don't know how to derive Traversable instances automatically.

So the answer - many instances wouldn't compile and DeriveTraversable wouldn't work.

2017-05-03 12:56 GMT+03:00 Jonathon Delgado : OK, I understand why Traversable is useful here - thank you Chris and Dmitry!

The next question is why Traversable requires Foldable. I looked at the

traverse can fuse traverse f = List.foldr cons_f (pure []) where cons_f x ys = (:) <$> f x <*> ys It uses List.foldr. Many other instances do the same. source, and couldn't see where Foldable is being used, other than as a constraint on Traversable. To put the question differently, what would fail to compile if this constraint was removed?

...

From: Dmitry Olshansky Sent: 03 May 2017 09:53 To: Jonathon Delgado

Cc: haskell-cafe@haskell.org Subject: 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

...

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

Chris Smith 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

Haskell-Cafe Info Page mail.haskell.org This mailing list is for the discussion of topics related to Haskell. The volume may at times be high, as the scope is broader than the main Haskell mailing list.

...

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Hmm, you are right! My objections were invalid. So I don't know an answer... Really, why we have this constraint? The same question is about Functor. 2017-05-03 17:43 GMT+03:00 Richard Eisenberg :

To me, the fact that DeriveTraversable requires a Foldable instance is not an argument saying that Foldable needs to be a superclass of Traversable. It just restricts the usefulness of DeriveTraversable. We could always advertise that DeriveTraversable works only on types with Foldable instances -- no need to bake this restriction into the superclass constraints of Traversable.

To be clear, I'm *not* arguing that Foldable shouldn't be a superclass of Traversable. I'm not knowledgeable enough on the specifics to make that claim. I *am* arguing that support for DeriveTraversable doesn't seem to a great argument for putting Foldable as a superclass of Traversable, though.

Richard

On May 3, 2017, at 10:38 AM, Dmitry Olshansky wrote:

Traversable instances (for "singletons" or not) can be implemented _uniformly_ using Functor and Foldable instances. And they are implemented in such way when you derive Traversable. So we need Foldable constraint for the class.

E.g. {-# LANGUAGE DeriveFunctor, DeriveFoldable,DeriveTraversable #-}

...

data D a = D { f1 :: [a], f2 :: (String, a) } deriving (Show, Eq, Functor, Foldable, Traversable) sum $ D [1..3] ("test",4) 10 mapM_ print $ sequenceA $ D [[1,2],[3,4]] ("test",[5,6]) D {f1 = [1,3], f2 = ("test",5)} D {f1 = [1,3], f2 = ("test",6)} D {f1 = [1,4], f2 = ("test",5)} D {f1 = [1,4], f2 = ("test",6)} D {f1 = [2,3], f2 = ("test",5)} D {f1 = [2,3], f2 = ("test",6)} D {f1 = [2,4], f2 = ("test",5)} D {f1 = [2,4], f2 = ("test",6)}

2017-05-03 14:34 GMT+03:00 Jonathon Delgado :

...

Interesting, that's not the one linked to from Dmitry's code.

In any case, is this correct?

1) Traversable is useful for all containers, including ones which can only hold a single value, such as (,) a. 2) The traversable definition for containers which can hold multiple versions requires Foldable. 3) So Traversable has to depend on Foldable. 4) So singleton containers also have to implement Foldable, even when it doesn't really make sense to do so.

Is there some kind of refactoring which would "fix" this, other than two unrelated Traversable classes? I understand that it might be impractical to refactor a widely-used standard library, but I would be interested in how such a situation could be avoided when starting from scratch.

From: Haskell-Cafe on behalf of Tony Morris Sent: 03 May 2017 11:21 To: haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] Foldable for (,)

https://i.imgur.com/A2enuhq.png

...

List.foldr has signature (a -> b -> b) -> b -> [a] -> b, i.e. an actual

On 03/05/17 21:17, Jonathon Delgado wrote: list? How is this effected by the Foldable constraint?

...

From: Dmitry Olshansky Sent: 03 May 2017 10:47 To: Jonathon Delgado Cc: haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] Foldable for (,)

Look how instance for List is defined.

instance Traversable [] where {-# INLINE traverse #-} -- so that

...

Functions in all instances of class should have the same signatures. So we have to add Foldable constraint to the class. Of cause we can implement 'foldr' internaly in 'traverse' if needed (as well as fmap). But this is not so good and more important that in this case we don't know how to derive Traversable instances automatically.

So the answer - many instances wouldn't compile and DeriveTraversable wouldn't work.

2017-05-03 12:56 GMT+03:00 Jonathon Delgado : OK, I understand why Traversable is useful here - thank you Chris and Dmitry!

The next question is why Traversable requires Foldable. I looked at the

traverse can fuse traverse f = List.foldr cons_f (pure []) where cons_f x ys = (:) <$> f x <*> ys It uses List.foldr. Many other instances do the same. source, and couldn't see where Foldable is being used, other than as a constraint on Traversable. To put the question differently, what would fail to compile if this constraint was removed?

...

From: Dmitry Olshansky Sent: 03 May 2017 09:53 To: Jonathon Delgado

Cc: haskell-cafe@haskell.org Subject: 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

...

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

Chris Smith 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

Haskell-Cafe Info Page mail.haskell.org This mailing list is for the discussion of topics related to Haskell. The volume may at times be high, as the scope is broader than the main Haskell mailing list.

...

...

Only members subscribed via the mailman list are allowed to post.

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Only members subscribed via the mailman list are allowed to post.

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It's a nice challenge to come up with datatypes that could be Traversable but not Foldable. First of all, you can think of Traversable as an extension of Functor. Instead of only mapping over the structure with a pure function, you can now also map over it with a "functorial function" (a -> f b) (I don't know of a better name, so I'm inventing one.) What that means is to have a Traversable structure, you need to have the ability to touch every element. If you can touch every element, why would you not be able to feed them all into a folding function? Only a few possible anwers come to mind for me: 1. There might be no obvious ordering. But in practice, you just invent one. 2. The structure might be infinite, so folding would never succeed. But laziness ftw. 3. Generalizing the previous answer, the structure might be computed ad-hoc and contain fixed points. That alone might not suffice, but it sounds like a more interesting starting point. 4. Generalizing a bit differently, the structure might be computed ad-hoc… because it's IO. IO surely can't be Foldable. But could it be Traversable? Well… no actually, because order is important. So can you come up with a structure that is computed ad-hoc, that contains fixed points, and where ordering is unimportant? Good luck. Cheers. On 2017-05-03 11:56, Jonathon Delgado wrote:

OK, I understand why Traversable is useful here - thank you Chris and Dmitry!

The next question is why Traversable requires Foldable. I looked at the source, and couldn't see where Foldable is being used, other than as a constraint on Traversable. To put the question differently, what would fail to compile if this constraint was removed?

From: Dmitry Olshansky Sent: 03 May 2017 09:53 To: Jonathon Delgado Cc: haskell-cafe@haskell.org Subject: 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.

_______________________________________________ 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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What we have here is a feature interaction problem. There's a combination of features (Foldable, Traversable, ((,) a), &c) each of which is sensible and useful, but which together give people a nasty surprise. We also have the combination of a language which is getting a lot of attention in terms of expressiveness, mathematical power, consistency and all manner of goodness BUT less attention with respect to beginners and bears of very little brain like me. If I write something that evaluates to length (2,3), it will quite definitely be unintentional. I'm reminded of a feature that Emacs used to have (and maybe still does): there were lots of commands that beginners were much more likely to type by accident than on purpose, so when you typed on, it would tell you something like "you just typed Ctrl-Meta-Explode, which does XYZ, are you sure you meant to do that?" and if you said yes, it remembered that and would let you use it again. That's one way to try to combine power for experts with safety for beginners. Again, if I remember correctly, PLT Scheme (now Racket) had similar support for beginners, where you could choose different language levels so that beginners didn't have to worry about major or even minor arcana. What I'm wondering here is if there's room in the Haskell world for something similar? Perhaps some sort of Profile: <profile name> pragma, where profiles can say things like "tuples are not Foldable". Does everyone have "beginner" days, or is it just me?

It seems that Traversable is doing two things: 1) Changing the shape of a data structure. 2) Folding over the contents of a data structure. Traversable requires Foldable to enable 2, but Traversable is also applied to types such as (,) where only 1 is relevant. If this is correct, follow-up questions would be: 1) For educational purposes, could these concerns be split without substantial drawback? 2) For practical purposes, could this be done without breaking a lot of existing code?

Hi Jonathon, Traversable instances do not change the shape of a data structure! On the contrary, the whole purpose of traverse is to combine effectful computations, stored in a structure without changing its shape! While Foldable folds over the contents of a data structure (via Monoid), Traversable allows you to fold over the effects attached to the contents of that structure (via Applicative). It turns out that by using Const applicative functor you can use traverse to perform foldMap! Thus every Traversable is trivially a Foldable (see foldMapDefault). Similarly, every Traversable is trivially a Functor if you use Identity applicative functor (see fmapDefault). This is why Traversable has those Functor and Foldable constraints, not because it relies on fmap or foldMap. Kind regards, Nick On Thu, 4 May 2017 at 10:49 Jonathon Delgado wrote:

It seems that Traversable is doing two things:

1) Changing the shape of a data structure. 2) Folding over the contents of a data structure.

Traversable requires Foldable to enable 2, but Traversable is also applied to types such as (,) where only 1 is relevant.

If this is correct, follow-up questions would be:

1) For educational purposes, could these concerns be split without substantial drawback? 2) For practical purposes, could this be done without breaking a lot of existing code? _______________________________________________ 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.

An alternate question might be, why is length a class member and not simply a function? The instance itself is merely documentation of a mathematical fact; it's only a few members of that instance (which are never part of its minimum definition!) that seem confusing. I'm sure the answer to that involves optimizations available for particular structures, case in point for tuples: length = const 1 But it's worth understanding it in detail, probably. On Wed, May 3, 2017 at 1:51 AM, Chris Smith wrote:

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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On May 3, 2017 at 11:36:00 AM, Michael Orlitzky (michael@orlitzky.com) wrote:

On 05/03/2017 04:44 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?

`const 42` is an equally-reasonable implementation to me. If the output is nonsense, who cares what particular nonsense it is?

These are all objectively stupid results:

By analogy, ask the same question about, say, PHP.

Question: why is "foo" == TRUE in PHP? Answer: it makes perfect sense, once you understand blah blah herp derp.

No, it's stupid, and your reasoning is stupid if you can justify it.

We’ve had some discussions about civility lately. This is also about having a productive discussion as well, I think. Comments like the above don’t meaningfully contribute to the discussion, and while they call “results” “objectively stupid” and not people, they tend to raise the temperature of a discussion in such a way that the latter seems only a few more flamewar posts away. I know people have deeply held opinions on things, but simply calling things “objectively stupid” repeatedly and loudly isn’t going to change any of those opinions. We know it won’t do that, because it hasn’t done that for the last two (?) years already. What we know it does, after experiencing it for the past two-ish years, is just make everyone sick of the mailinglists and increasingly testy towards one another, which is an outcome I think nobody wants. Can we stop with this, please? —Gershom

On Wed, May 3, 2017 at 1:11 PM, Andrey Sverdlichenko wrote:

people who don't like absurd semantic

Absurd like topology? Like category theory? Heck, like negative numbers? -- brandon s allbery kf8nh sine nomine associates allbery.b@gmail.com ballbery@sinenomine.net unix, openafs, kerberos, infrastructure, xmonad http://sinenomine.net

On 2017-05-03 19:11, Andrey Sverdlichenko wrote:

This is probably twentieth time this question arises. Discussion quickly boils down to people liking convenient syntax sugar Foldable instance gives (traverse pair, etc), versus people who don't like absurd semantic which becomes correct with such instances.

Language like "absurd semantic" perpetuates the negativity/confrontation that Gershom was talking about. Please try to avoid such language. (It's absolutely fine to disagree, but phrasing matters.) Regards,