Re: [Haskell] A puzzle and an annoying feature

Ketil Malde <ketil+haskell@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)? -kzm -- If I haven't seen further, it is by standing in the footprints of giants