5 Jul
2012
5 Jul
'12
1:17 p.m.
Thanks Francesco. And I did verify that ExplicitForAll does in fact allow Rank 1 Types in functions like the following ... f :: (forall a. a -> a) -- Rick On Thu, 2012-07-05 at 16:28 +0100, Francesco Mazzoli wrote:
At Thu, 05 Jul 2012 11:18:00 -0400, rickmurphy wrote:
data T = TC (forall a b. a -> b -> a)
The type of `TC' will be `(forall a b. a -> b -> a) -> T', a Rank-2 type.
-- Francesco * Often in error, never in doubt
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