[Haskell-cafe] ANN: package Boolean: Generalized booleans

[Conal Elliott]

Does this code compile?  I'd expect that

  instance Bool (Js JsBool) (Js r) where

violates the fundep, since it applies for *all* values of r, not just to
one.

   - Conal

On Tue, Jun 30, 2009 at 8:53 AM, Sebastiaan Visser <sfvisser at cs.uu.nl>wrote:

> On Jun 30, 2009, at 5:24 PM, Conal Elliott wrote:
>
>> Hi Sebastiaan,
>>
>> I like your extensions of generalized booleans to other common Haskell
>> types!
>>
>> I also prefer using type families to fundeps.  In this case I didn't
>> because of some awkwardness with vector operations, but I'm going to try
>> again.
>>
>> I'm confused about your particular fundep choice.  For instance,
>>
>> class Bool f r | f -> r where
>>  bool  :: r -> r -> f -> r
>>  false :: f
>>  true  :: f
>>
>> Do you *really* mean that the boolean type f determines the value type r?
>>
>
> Yes, that is really what I mean. This can be used to enforce that the
> return value of elimination can be restricted by the boolean type. This is
> especially useful when using GADTs to encode your domain language.
>
> For example, take this simple JavaScript language:
>  data Js a where
>    Prim :: String -> Js a                -- Primitive embedding.
>    App :: Js (a -> b) -> Js a -> Js b    -- Function application.
>
>  data JsBool
>
> Now the functional dependencies can be used to enforce that eliminating
> booleans in the Js domain always returns a value in the Js domain:
>  instance Bool (Js JsBool) (Js r) where
>    bool f t c = Prim "(function ifthenelse (f, t, c) c ? t : f)" `App` f
> `App` t `App` c
>    true  = Prim "true"
>    false = Prim "false"
> Getting rid of this fundep and using type families will probably be a lot
> more intuitive.
>
> Any suggestions on how to enforce elimination to be able to go from `Js
> JsBool -> Js r' using other techniques?
>
>> Regards,   - Conal
>>
>> ...
>>
>
> --
> Sebastiaan Visser
>
>
>
>

[]
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