[Haskell] A puzzle and an annoying feature
ketil+haskell at ii.uib.no
Thu Nov 25 01:01:51 EST 2004
Ketil Malde <ketil+haskell at ii.uib.no> writes:
> Did you perhaps mean:
> g :: a -> a = \x -> x
> which has type () -> () ?
Or maybe the difference between:
g :: Num a => a -> a
g = \x -> 1
(which gives the specified type) and
g' :: forall a . Num a => a -> a
g' = \x -> 1
(which gives Integer -> Integer)?
If I haven't seen further, it is by standing in the footprints of giants
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