Hi Michael On 28 Jul 2008, at 07:50, Michael Karcher wrote: David wrote (with patch applied):
newtype Backward f a = Backward { runBackward :: f a } deriving Functor
instance Applicative f => Applicative (Backward f) where pure = Backward . pure (Backward f) <*> (Backward a) = Backward (a <**> f)
You said:
According to the haddock of Control.Applicative, this line is semantically equivalent to Backward f <*> Backward a = Backward (f <*> a)
I'm not saying the haddock is entirely clear, but it certainly doesn't necessitate the interpretation you're making.
So I don't see the point of this "transformer". It seems to do nothing, just as newtype Backward f a = Backward { runBackward :: f a } deriving (Functor, Applicative) would also do.
Appearances can be deceptive, so why not actually try it? *Backward> traverse print ["bong", "bing"] "bong" "bing" [(),()] *Backward> runBackward $ traverse (Backward . print) ["bong", "bing"] "bing" "bong" [(),()] (<*>) does the function's effects before the argument's; (<**>) does the argument's effects before the function's. In contrast with monads, the applicative interface does not offer the ability to make the choice of one computation depend on the value of another (in Lindley/Wadler/Yallop terminology, applicative functors are "oblivious"). One consequence is that it's easy to reverse the order of computation. All the best Conor