On 18 March 2011 14:54, Tyson Whitehead
map f . map g = map (f . g)
should be derivable from the type of map "(a -> b) -> [a] -> [b]"
I don't think this is true as stated. Consider: map f [] = [] map f [x] = [f x] map f (x:y:zs) = f x : map zs Your proposed property is violated though we have the appropriate type. IIRC I think what is true is that map f . map' g = map' (f . g) For any map' :: (a -> b) -> [a] -> [b]
class Seq a where seq :: a -> b -> b
AFAIK Haskell used to have this but it was removed because a polymorphic seq is just so damn convenient! Furthermore, this solution would not help unless you did not have a Seq (a -> b) instance, but such an instance could be necessary to avoid some space leaks, since you can write stuff like: case holds_on_to_lots_of_memory of True -> \x -> x-1 False -> \x -> x+1 So there are eminently practical reasons for polymorphic seq. Cheers, Max