On 2/28/11 6:01 AM, Yves Parès wrote:
takeC :: Int -> Compoz a b -> (exists c. Compoz a c)
dropC :: Int -> Compoz a b -> (exists c. Compoz c b)
What does 'exists' means? To create a rank-2 type can't you use:
takeC :: Int -> Compoz a b -> (forall c. Compoz a c)
??
For any A and T,
foo :: A -> (forall b. T b)
is identical to
foo :: forall b. (A -> T b)
More technically, they're isomorphic--- in System F or any other gritty
language that makes you explicitly pass types around. Since Haskell
handles type passing implicitly, the isomorphism looks like identity.
--
Live well,
~wren
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