On Mon, Feb 28, 2005 at 11:10:40PM -0500, Jim Apple wrote:
Jon Fairbairn wrote:
If you allow quantification over higher
kinds, you can do something like this:
d f = f . f
d:: a::*, b::**.(b a a) b (b a) a
What's the problem with
d :: (forall c . b c - c) - b (b a) - a
d f
It is really too bad the 'middle' version does not work, ie
John Fairbarn's version
d1 :: (forall c . b c - c) - b (b a) - a
d1 f = f . f
John Meacham's version (dual (?))
d2 :: (forall c . c - b c) - a - b (b a)
d2 f = f . f
Or something in the middle
d3 :: forall e a b . (forall c . e
Actually none of these seem to work:
{-# OPTIONS -fglasgow-exts #-}
module Main where
main :: IO ()
main = putStrLn OK
d :: (forall c . b c - c) - b (b a) - a
d f = f . f
t0 = d id
t1 = d head
t2 = d fst
Load this into GHCI and you get:
Test.hs:11:7:
Couldn't match the rigid variable `c'
On 2005-02-28 at 23:10EST Jim Apple wrote:
Jon Fairbairn wrote:
If you allow quantification over higher
kinds, you can do something like this:
d f = f . f
d:: a::*, b::**.(b a a) b (b a) a
What's the problem with
d :: (forall c . b c - c) - b (b a) - a
d f = f . f
Jon Fairbairn wrote:
If you allow quantification over higher
kinds, you can do something like this:
d f = f . f
d:: a::*, b::**.(b a a) b (b a) a
What's the problem with
d :: (forall c . b c - c) - b (b a) - a
d f = f . f
to which ghci gives the type
d :: forall a b. (forall c. b c - c) - b
Jon Fairbairn wrote:
If you allow quantification over higher
kinds, you can do something like this:
d f = f . f
d:: a::*, b::**.(b a a) b (b a) a
What's the problem with
d :: (forall c . b c - c) - b (b a) - a
d f = f . f
to which ghci gives the type
d :: forall a b. (forall c. b c - c) -
