[Haskell-cafe] Get number of fields for non-record ADTs using Data type class?
Li-yao Xia
lysxia at gmail.com
Mon Oct 29 21:52:31 UTC 2018
On 10/29/18 4:42 PM, Artem Pelenitsyn wrote:
> I don't think there is a point in looking for
> GHC.Generics-based solution, as Data.Data is the exact match for this
> kind of problem.
Although GHC.Generics has its shortcomings (usage complexity and compile
times), I still find it worthwhile to advocate as a statically-typed
alternative to the dynamically-typed Data.Data for many problems,
including this one.
Using the one-liner library (which is built around GHC.Generics), the
equivalent line of code is:
getSum . gfoldMap @AnyType (const 1) :: T -> Int
Data.Data is more visible mainly because it comes with a lot of
functionality baked into the standard library, whereas GHC.Generics
provides only a minimal interface and we have to find everything else in
separate packages. However, there is no fundamental reason why one is a
better fit than the other for the task of counting constructor fields.
Li-yao
>
> On Mon, 29 Oct 2018 at 16:35 Li-yao Xia <lysxia at gmail.com
> <mailto:lysxia at gmail.com>> wrote:
>
> This maps every field to 1, and folds them together using (+):
>
> Data.Data.gmapQl (+) 0 (const 1) :: T -> Int
>
> (There has to be a similarly easy solution using GHC.Generics instead
> but I can't think of one...)
>
> Li-yao
>
> On 10/29/18 2:56 PM, Markus Läll wrote:
> > Dear list,
> >
> > Is it possible te get the number of fields for data constructors
> for a
> > plain ADT, i.e something with no record fields? E.g for
> >
> > data T = A Int Double | B String (Maybe String)
> >
> > it would give 2 for both `A` and `B`.
> >
> > For a record it's possible using the `constrFields` function from
> Data.Data.
> >
> > I was trying to follow this tutorial by Christopher Done
> > https://chrisdone.com/posts/data-typeable, and I feel that it
> must be
> > possible somehow to get these numbers with the gmap*/gfold*
> functions,
> > but the use of them is over my head at the moment.
> >
> >
> > Best,
> >
> >
> >
> > --
> > Markus Läll
> >
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