[Haskell-cafe] Abstraction in data types

[Lyndon Maydwell]

Well you need to define a new datatype to make a hertrogenous list, so
I don't think there's any real way you can get around people doing
that...

On Thu, Mar 18, 2010 at 1:27 PM, Darrin Chandler
<dwchandler at stilyagin.com> wrote:
> On Thu, Mar 18, 2010 at 01:06:25PM +0800, Lyndon Maydwell wrote:
>> You could probably also use a typeclass for pointy things rather than
>> a data type, this would then require you to use existential
>> quantification to construct a hetrogenous list.
>>
>> For example:
>>
>> Class Point where
>>     getCartesian :: ...
>>     getPolar :: ...
>>
>> data Shape = Point p => ... | Polygon [p]
>>
>> Correct me if this is wrong though :-)
>
> So in "normal" use Polygon list would be homogeneous, but could be made
> heterogeneous with effort? If I have that right it's closer, but I'd
> love to have the compiler whine if someone tried to mix them.
>
>> On Thu, Mar 18, 2010 at 12:56 PM, Alexander Solla <ajs at 2piix.com> wrote:
>> > I wrote this to Darrin, but didn't CC cafe:
>> > On Mar 17, 2010, at 9:20 PM, Darrin Chandler wrote:
>> >
>> > type Cartesian_coord = Float
>> >
>> > type Latitude  = Float
>> > type Longitude = Float
>> >
>> > data Point = Cartesian (Cartesian_coord, Cartesian_coord)
>> > | Spherical (Latitude, Longitude)
>> >
>> > type Center = Point
>> > type Radius = Float
>> >
>> > data Shape = Circle Center Radius
>> > | Polygon [Point]
>> >
>> > This obviously stinks since a Polygon could contain mixed Cartesian and
>> > Spherical points. Polygon needs to be one or the other, but not mixed.
>> >
>> > My suggestion would be to use an alternate representation of "spherical"
>> > points in terms of polar coordinates, and then to normalize and mix at will:
>> > type Theta = Float
>> > type Radius = Float
>> > data Point = Cartesian (Cartesian_coord, Cartesian_coord)
>> >            | Polar   (Theta, Radius)
>> > normalize_point :: Point -> Point
>> > normalize_point Cartesian x y = Cartesian x y
>> > normalize_point Polar t r = Cartesian x y where x = r * cos t; y = r * sin
>> > t;
>> > It really depends on what you want to do with your points.  If you want to
>> > do linear algebra, you might want your points to depend on a basis, for
>> > example.  But your "spherical" points don't really form a basis in
>> > three-space, or even over all of two-space.
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>
> --
> Darrin Chandler            |  Phoenix BSD User Group  |  MetaBUG
> dwchandler at stilyagin.com   |  http://phxbug.org/      |  http://metabug.org/
> http://www.stilyagin.com/  |  Daemons in the Desert   |  Global BUG Federation
>