You could do something like this, but admittedly it appears slightly clunky: newtype Baz f a = Baz (Foo f => BazInner f a) data BazInner f a = BazInner { baz :: f a, baz2 :: f a } instance Foo (Baz f) where foo a = Baz (let b = foo a in BazInner b b) Cheers, Daniel On 2011-June-09 Thursday 15:25:40 Guy wrote:

Can this be extended to records, without redundant repetition?

data Baz f a = Baz {baz :: Foo f => f a, baz2 :: Foo f => f a}

The type constraint for baz2 adds no information, as it's the same f as baz, but I can't leave it out.

----- Original Message -----

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From: Daniel Schüssler To: haskell-cafe@haskell.org Cc: Guy Sent: Thursday, 9 June 2011, 2:06 Subject: Re: [Haskell-cafe] Type Constraints on Data Constructors

Hello,

you might be thinking of this type?

{-# LANGUAGE Rank2Types #-}

class Foo f where foo :: a -> f a

data Baz f a = Baz (forall f. Foo f => f a)

instance Foo (Baz f) where foo a = Baz (foo a)

Maybe the difference between Bar and Baz ist best explained by writing it with an explicit class dictionary for Foo:

{-# LANGUAGE Rank2Types #-}

data FooDict f = FooDict { foo :: forall a. a -> f a }

data Bar f a = Bar (FooDict f) (f a)

data Baz f a = Baz (FooDict f -> f a)

fooDict_Baz :: FooDict (Baz f) fooDict_Baz = FooDict (\a -> Baz (\d -> foo d a))

-- fooDict_Bar :: FooDict (Bar f) -- fooDict_Bar = FooDict (\a -> Bar ? ?) -- Doesn't work - you'd have to create a 'FooDict f' and a 'f a' out of just an 'a'

Cheers, Daniel

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