So are you saying data Shish2 a = Holder2 a | Onion (Shish a) | Lamb (Shish a) | Tomato (Shish a) but then I'm having trouble with meal6 = (Onion (Tomato (Lamb (Holder2 (Fork))))) Couldn't match expected type `Shish a1' with actual type `Shish2 a0' * In the first argument of `Tomato', namely `(Lamb (Holder2 (Fork)))' In the first argument of `Onion', namely `(Tomato (Lamb (Holder2 (Fork))))' In the expression: (Onion (Tomato (Lamb (Holder2 (Fork))) On Fri, Mar 12, 2021 at 4:08 PM Tikhon Jelvis wrote:

Onion of 'a shish is equivalent to Onion (Shish a) in Haskell rather than Onion a (Shish a).

The latter version in Haskell creates a constructor with two arguments, something like Onion of ('a * 'a shish) would in SML. (Or, at least, OCaml—not 100% familiar with SML syntax myself!)

On Fri, Mar 12, 2021, 14:03 Galaxy Being wrote:

...

Hello,

This is my first post here, and it's an odd one, I'll admit. Basically, I'm trying to translate the material in *The Little MLer *to Haskell, the TLMLer being an excellent types workout. So in SML I have this

datatype 'a shish = Bottom of 'a | Onion of 'a shish | Lamb of 'a shish | Tomato of 'a shish

and this

datatype rod = Dagger | Fork | Sword

and then this SML function

fun is_veggie (Bottom (x)) = true | is_veggie (Onion (x)) = is_veggie (x) | is_veggie (Lamb (x)) = false | is_veggie (Tomato (x)) = is_veggie (x)

which has no problem handling tis

is_veggie (Onion(Tomato(Bottom(Dagger))))

Now, in Haskell I've translated this (with minor alterations) to

data Shish a = Holder a | Onion a (Shish a) | Lamb a (Shish a) | Tomato a (Shish a) data Rod = Dagger | Fork | Sword

However, in Haskell these two expressions are different things entirely

meal4 = Tomato Dagger (Onion Fork (Lamb Spear (Holder Fork))) meal5 = (Tomato (Onion (Lamb (Holder Fork))))

Here's my attempt at handling meal4 with a Haskell isVeggie

isVeggie (Holder (sh)) = True isVeggie (Onion sh (sk)) = isVeggie sk isVeggie (Tomato sh (sk)) = isVeggie sk isVeggie (Lamb sh (sk)) = False

This works for meal4, but not for meal5. And yet in the SML world their is_veggie handles (Onion(Tomato(Bottom(Dagger)))) just fine. TLMLer says

Onion (Tomato (Bottom (Dagger)))

belongs to the type rod shish, while in Haskell

Onion (Tomato (Holder (Dagger)))

is a bizarre nested beast due to the fact that the data constructor variable of Onion is Tomato (Holder (Dagger)) etc. etc.

Can a single Haskell version of isVeggie handle both meal4 and meal5? No? I thought so. But then how would a separate Haskell version of isVeggie handle meal5 -- or is it just too weird? Also, but not critical, how could the Haskell isVeggie be done with guards, i.e., just like a consed list is matched on (x:xs) in the recursion case? I can see that 1:2:3:[] and Onion (Tomato (Bottom (Dagger))) are both conses, but the latter I don't know how to break out into head and tail for a guard case where the individual food items were not mentioned explicitly. IOW, this doesn't work

isVeggieKebab :: Shish -> Bool isVeggieKebab Holder (sk) = True isVeggieKebab (shkb (sk)) | (shkb == Onion) || (shkb == Tomato) = isVeggieKebab sk | otherwise = False

I'm feeling some doom and gloom about this project. Right at the start this seems to be an insurmountable difference between SML and Haskell type systems. Or I simply don't understand something fundamental here.

LB

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So embarrassing. Thanks. Anyway, this is what works data Shish2 a = Holder2 a | Onion2 (Shish2 a) | Lamb2 (Shish2 a) | Tomato2 (Shish2 a) whatHolder2 (Holder2 (sh)) = sh whatHolder2 (Onion2 (sk)) = whatHolder2 sk whatHolder2 (Tomato2 (sk)) = whatHolder2 sk whatHolder2 (Lamb2 (sk)) = whatHolder2 sk meal6 = (Onion2 (Tomato2 (Lamb2 (Holder2 (Fork)))))

whatHolder2 meal6 Fork

However, I'm still wondering how to have an abstracted (x:xs) - like pattern to collapse all the ingredients, i.e., whatHolder2 (Holder2 (sh)) = sh whatHolder2 (shish2-head (shish2-tail)) = whatHolder2 shish2-tail On Fri, Mar 12, 2021 at 4:44 PM Bob Ippolito wrote:

I think the problem is that you have a typo such that this Shish2 is dependent on some other data type named Shish. If you change the type name to Shish (or change its definition to use Shish2) it works fine.

data Shish a = Holder2 a | Onion (Shish a) | Lamb (Shish a) | Tomato (Shish a) deriving Show data Rod = Dagger | Fork | Sword deriving Show

meal6 = (Onion (Tomato (Lamb (Holder2 (Fork)))))

On Fri, Mar 12, 2021 at 2:23 PM Galaxy Being wrote:

...

So are you saying

data Shish2 a = Holder2 a | Onion (Shish a) | Lamb (Shish a) | Tomato (Shish a)

but then I'm having trouble with

meal6 = (Onion (Tomato (Lamb (Holder2 (Fork)))))

Couldn't match expected type `Shish a1' with actual type `Shish2 a0' * In the first argument of `Tomato', namely `(Lamb (Holder2 (Fork)))' In the first argument of `Onion', namely `(Tomato (Lamb (Holder2 (Fork))))' In the expression: (Onion (Tomato (Lamb (Holder2 (Fork)))

On Fri, Mar 12, 2021 at 4:08 PM Tikhon Jelvis wrote:

...

Onion of 'a shish is equivalent to Onion (Shish a) in Haskell rather than Onion a (Shish a).

The latter version in Haskell creates a constructor with two arguments, something like Onion of ('a * 'a shish) would in SML. (Or, at least, OCaml—not 100% familiar with SML syntax myself!)

On Fri, Mar 12, 2021, 14:03 Galaxy Being wrote:

...

Hello,

This is my first post here, and it's an odd one, I'll admit. Basically, I'm trying to translate the material in *The Little MLer *to Haskell, the TLMLer being an excellent types workout. So in SML I have this

datatype 'a shish = Bottom of 'a | Onion of 'a shish | Lamb of 'a shish | Tomato of 'a shish

and this

datatype rod = Dagger | Fork | Sword

and then this SML function

fun is_veggie (Bottom (x)) = true | is_veggie (Onion (x)) = is_veggie (x) | is_veggie (Lamb (x)) = false | is_veggie (Tomato (x)) = is_veggie (x)

which has no problem handling tis

is_veggie (Onion(Tomato(Bottom(Dagger))))

Now, in Haskell I've translated this (with minor alterations) to

data Shish a = Holder a | Onion a (Shish a) | Lamb a (Shish a) | Tomato a (Shish a) data Rod = Dagger | Fork | Sword

However, in Haskell these two expressions are different things entirely

meal4 = Tomato Dagger (Onion Fork (Lamb Spear (Holder Fork))) meal5 = (Tomato (Onion (Lamb (Holder Fork))))

Here's my attempt at handling meal4 with a Haskell isVeggie

isVeggie (Holder (sh)) = True isVeggie (Onion sh (sk)) = isVeggie sk isVeggie (Tomato sh (sk)) = isVeggie sk isVeggie (Lamb sh (sk)) = False

This works for meal4, but not for meal5. And yet in the SML world their is_veggie handles (Onion(Tomato(Bottom(Dagger)))) just fine. TLMLer says

Onion (Tomato (Bottom (Dagger)))

belongs to the type rod shish, while in Haskell

Onion (Tomato (Holder (Dagger)))

is a bizarre nested beast due to the fact that the data constructor variable of Onion is Tomato (Holder (Dagger)) etc. etc.

Can a single Haskell version of isVeggie handle both meal4 and meal5? No? I thought so. But then how would a separate Haskell version of isVeggie handle meal5 -- or is it just too weird? Also, but not critical, how could the Haskell isVeggie be done with guards, i.e., just like a consed list is matched on (x:xs) in the recursion case? I can see that 1:2:3:[] and Onion (Tomato (Bottom (Dagger))) are both conses, but the latter I don't know how to break out into head and tail for a guard case where the individual food items were not mentioned explicitly. IOW, this doesn't work

isVeggieKebab :: Shish -> Bool isVeggieKebab Holder (sk) = True isVeggieKebab (shkb (sk)) | (shkb == Onion) || (shkb == Tomato) = isVeggieKebab sk | otherwise = False

I'm feeling some doom and gloom about this project. Right at the start this seems to be an insurmountable difference between SML and Haskell type systems. Or I simply don't understand something fundamental here.

LB

_______________________________________________ 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.