[Haskell] Re: Type of y f = f . f

[Keean Schupke]

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' against `b c'
      `c' is bound by the polymorphic type `forall c. b c -> c' at 
Test.hs:11:5-8
      Expected type: b c -> c
      Inferred type: b c -> b c
    In the first argument of `d', namely `id'
    In the definition of `t0': t0 = d id

Test.hs:13:5:
    Inferred type is less polymorphic than expected
      Quantified type variable `c' escapes
      It is mentioned in the environment:
        t2 :: (c, (c, a)) -> a (bound at Test.hs:13:0)
    In the first argument of `d', namely `fst'
    In the definition of `t2': t2 = d fst
Failed, modules loaded: none.

Keean.


Jacques Carette wrote:

>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 c -> b c) -> (e a) -> (b a)
>>d3 f = f . f 
>>    
>>
>
>but ghci -fglasgow-exts does not like it :-(
>
>Jacques
>
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