Hi,
Can anyone explain to me how hugs manages to derive that
f x y z = y (y z) x
is of type
f :: a -> ((a -> b) -> a -> b) -> (a -> b) -> b
Many thanks and a happy new year to all!
Lee
_
Stay in touch with absent friends - get M
Hi,
I've run into a small problem whilst doing some manual type checking, to see
if I could match the results given by hugs
prelude> :t foldr filter
foldr filter :: [a] -> [a -> Bool] -> [a]
-- This was fine and was the same as my answer, so I tested it with
prelude> foldr filter [1,2,3,4] [even
