[Haskell-beginners] Re: Classes in Polymorphic Constructors
voldermort at hotmail.com
Mon Aug 23 15:17:18 EDT 2010
On 23/08/2010 20:28, Brandon S Allbery KF8NH wrote:
> On 8/23/10 12:12 , John Smith wrote:
>> On 23/08/2010 18:32, Ertugrul Soeylemez wrote:
>>> John Smith<voldermort at hotmail.com> wrote:
>>>> Why can polymorphic data constructors not be instantiated as a class?
>>>> (not sure if I used the right terminology there)
>>>> For example,
>>>> a = [1,True]::[Show]
>>>> b = "foo":a
>>>> c = map show b
>>>> This would obviate the need for wrapper types
>>> Your type signature doesn't make sense. You are confusing type classes
>>> with types. Show is a type class.
>> My question was why doesn't the type system allow classes to be used in this
>> way. It's the equivalent of a base class or interface in OO.
> Typeclasses are not classes, and Haskell values are not objects. In OO
> languages, objects carry around dictionaries with full details of what they
> can do. Haskell values, on the other hand, don't actually carry around
> anything; what they can do is specified by the type. If the type is (Show a
> => [a]), that means the *only* thing you can do with an (a) is to call
> (show) on it.
> (1) Since this is almost certainly not what you intend,
Wouldn't this be a useful feature? For example, [Show] could contain a list of values for serialising to strings, and
ditto for any other typeclass. When you say "Since this is almost certainly not what you intend", is there something
else which could be intended by such code?
> Haskell insists you
> acknowledge that by specifying the type as ([forall a. Show a => a]), making
> it explicit that the scope of (a) is limited to inside the brackets ---
> meaning that, since anything wanting to use it has to go through the
> brackets, *all* it knows is that it can invoke (show) on it.
> (2) Again, values are not objects --- there is no method dictionary hidden
> inside of them. Typeclass dictionaries are carried *outside* of values, as
> hidden arguments to functions that are declared with types that are
> typeclass members. Which is why you can't say (a = [1,True] :: [forall a.
> Show a => a]): there's nowhere to put the dictionary in the value
> ([1,True]). This is why a wrapper is required; the wrapper provides a place
> for the hidden argument.
> Summary: If you want an OO language, use an OO language. Haskell isn't one.
> There are some outdated examples of OO extensions for Haskell that I don't
> think are maintained any more, notably "OOHaskell". They're not maintained
> due to lack of interest. If you want such a thing, you'll have to modernize
> it yourself.
I don't want OO, just for typeclasses to be usable as types in polymorphism, seeing as both are equally relevant as a
definition of constraints. (There's probably a better way to word that.)
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