Re: [Haskell-beginners] Equivalence of Inheritance

14 Dec 2010


      Ozgur pointed out this notition p@Man{}.  What I would like is something
like this.

type Man = Person@Man{}

type MensGroup = [Man]
*
-- Russ *


On Tue, Dec 14, 2010 at 1:45 PM, Michael Katelman  wrote:
...
Without going too exotic, I see two choices. Keeping the single type
Person makes what you are asking for, as far as I know, impossible
(dependent types). For human readability you could consider
type MensGroup = [Person]
type WomensGroup = [Person]
If you split Person into two types, Man and Woman, there are
repercussions for the aggregate group type
type Group = [Either Man Woman]
-Mike
On Tue, Dec 14, 2010 at 3:30 PM, Russ Abbott 
wrote:
...
Now that this is straightened out, I went back to what I was doing in the
first place and realized that I haven't solved my problem.
Given
data Person =
      Man {name :: String, age :: Int, prostateCondition :: Condition}
  | Woman {name :: String, age :: Int, ovaryCondition    :: Condition}
I'd like to define something like this.
type MensGroup = [Man]
Is there a way to do something like that?
-- Russ
On Tue, Dec 14, 2010 at 1:06 PM, Russ Abbott 
wrote:
...
That's good. (It's more or less the way I was doing it.)  What I wanted
to
...
avoid was this.
getGenderSpecificCondition (  Man _ _ cond) = cond
getGenderSpecificCondition (Woman _ _ cond) = cond
I know it seems like a small thing, but I would like to be able to write
it like this.
getGenderSpecificCondition p
   | p == (Man _ _ cond) = cond
   | p == (Woman _ _ cond) = cond
But that's not legal syntax.  A pattern can't appear in that context.
But
this does the job.
getGenderSpecificCondition :: Person -> Condition
getGenderSpecificCondition p
   | isMan p = prostateCondition p
   | isWoman p = ovaryCondition p
isMan (     Man _ _ cond) = True
isMan _ = False
isWoman (Woman _ _ cond) = True
isWoman _ = False
That works! prostateCondition and ovaryCondition are both defined on
Person. (I'm surprised to see that.)
*Person> Group [Man "Harry" 32 OK, Woman "Sally" 29 Good]
Harry(32, OK)
Sally(29, Good)
Also
*Person> prostateCondition (Woman "Sally" 29 Good)
*** Exception: No match in record selector prostateCondition
*Person> prostateCondition (Man "Harry" 29 Good)
Good
-- Russ
On Tue, Dec 14, 2010 at 12:31 PM, Michael Katelman 
wrote:
...
Perhaps this?
https://gist.github.com/741048
-Mike
On Tue, Dec 14, 2010 at 2:27 PM, Russ Abbott 
wrote:
...
What I'm after is a version of my example that compiles.  Can you
make
...
...
one?
-- Russ
On Tue, Dec 14, 2010 at 12:18 PM, Antoine Latter 
...
...
...
wrote:
...
Sorry, I really don't know enough about what you're after to attempt
that.
But you'll need to change you're signatures of the form:
> function :: Person -> Foo
to something of the form:
> function :: Person p => p -> Foo
Because again, a type class can not be used as a type.
Antoine
On Tue, Dec 14, 2010 at 2:12 PM, Russ Abbott 
...
...
...
...
wrote:
> What got fouled up is all the adjustments I had to make to the
other
> declarations.
> Can you complete the example so that it compiles using
>
> class Person p where ...
>
> I'd very much like to see an example that actually compiles.
>
> Thanks.
> -- Russ
>
> On Tue, Dec 14, 2010 at 11:58 AM, Antoine Latter
> 
> wrote:
>>
>> On Tue, Dec 14, 2010 at 1:52 PM, Russ Abbott
>> 
>> wrote:
>> > If gender is a field in a Person type, then a Person must have
>> > both
>> > an
>> > ovaryCondition and a prostateCondition.  That seems awkward.
>> > Regarding
>> >      class Person p where
>> > I started down that path but got completely fouled up.
>>
>> How did this get fouled up? Every class declaration must take
>> arguments - here, 'p' is the argument for the class.
>>
>> Thanks,
>> Antoine
>
>
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