You can do this, although you still need a datastructure that allows you to use the contained type: {-# LANGUAGE GADTs #-} {-# LANGUAGE StandaloneDeriving #-} {-# LANGUAGE TypeSynonymInstances #-} module Tutorial where data Gender = Male | Female deriving (Show) data Race = White | Black deriving (Show) type Age = Int data Answer a where Answer :: RacistAgistSexist a => a -> Answer a deriving instance Show w => Show (Answer w) data GenderRaceAge = Gender Gender | Race Race | Age Age class RacistAgistSexist a where genderRaceAge :: a -> GenderRaceAge instance RacistAgistSexist Gender where genderRaceAge = Gender instance RacistAgistSexist Race where genderRaceAge = Race instance RacistAgistSexist Age where genderRaceAge = Age -- You can use genderRaceAge to get a GenderRaceAge out of the contained type if you don't know 'a' On 15 May 2015 at 08:51, Tom Ellis < tom-lists-haskell-cafe-2013@jaguarpaw.co.uk> wrote:
On Fri, May 15, 2015 at 01:47:49AM -0500, Cody Goodman wrote:
How can I create Answers of type Gender, Race, or Age?
These should be possible:
λ> Answer Male λ> Answer White λ> Answer Black λ> Answer 28
Others such as using a string should not be possible:
λ> Answer "a string" -- should throw type error
It would probably help if you tell us why precisely you want this, and in particular why
data Answer = AnswerGender Gender | AnswerRace Race | AnswerAge Int
is not satisfactory.
Tom
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