Get a Haskell record's field names as a list of strings?

Question

Say I have the following:

data Rec = Rec {
    alpha :: Int,
    beta  :: Double,
    phi   :: Float 
} 

sample = Rec 1 2.3 4.5

I understand Template Haskell & the reify function can get me the record's field names. That is:

print $(f sample) --> ["alpha", "beta", "phi"]

There is also a claim that this can be done without Template Haskell. Can someone provide an example implementation for this can be accomplished?

Expecting

You can get the field names without template Haskell too. – Kenton

@augustss: How? Some Typable black magic? Either way, most uses one could put this information to would be a great fit for Template Haskell. – Georgy

@delnan You can either use Data.Data or you can just derive Show, show sample and do a little parsing of that string. – Kenton

I don't see many compelling reasons to actually do this. – Leena

Answer 1

It can be done with a Data (most GHC versions) or Generic (7.2.x and up) instance, which GHC can derive for you. Here's an example of how to dump record fields with the Data typeclass:

{-# LANGUAGE DeriveDataTypeable #-}

import Data.Data

data Rec = Rec {
    alpha :: Int,
    beta  :: Double,
    phi   :: Float 
}  deriving (Data, Typeable)

sample = Rec 1 2.3 4.5

main :: IO ()
main = print . constrFields . toConstr $ sample 

Planter

You can also avoid having to use an instance of the type (sample in this case) by using print . map constrFields . dataTypeConstrs . dataTypeOf $ (undefined :: Rec). This will generate a list of all the fields in all of the available constructors (and can of course be tailored to your liking) – Travelled