Re: [Haskell-cafe] Get number of fields for non-record ADTs using Data type class?

29 Oct 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 mailto:lysxia@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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