3 ms·

Typeclasses provide nothing but convenience there (see [1] for instance). The real power comes from the Algebraic Data Types which are intact. [1] http://www.h

• colanderman 13y ago
  I don't see how that mechanism provides subtyping (which typeclasses do)? Last I checked Haskell's structures weren't structurally typed.
  • tel 13y ago
    You could do it like in Agda, data ApplicativeI f = AI { pure :: forall a. a -> f a , ap :: forall a b. f (a -> b) -> f a -> f b , _FunctorI :: FunctorI f } data MonadI m = AI { bind :: forall a b. (a -> m b) -> m a -> m b , _ApplicativeI :: ApplicativeI m } return :: _MonadI m -> a -> m a return (MonadI { _ApplicativeI = ApplicativeI { pure = fn } }) = fn
    • colanderman 13y ago
      Almost, but that's not extensible: I can't declare some new typeclass Foo and then declare that every x s.t. Monad x is also Foo x without editing the definition of Monad.
      • tome 13y ago
        Instead of providing an instance declaration you provide a value of MonadI f -> FooI f.
        • tel 13y ago
          Exactly, and then that function is "proof" of the relationship.
• deleted 13y ago
  [deleted]
• tome 13y ago
  Personally I wouldn't want to do the instance plumbing by hand.