Alfredo Di Napoli
Thanks Brent, this should do the trick, although what I was asking was something more general:
For "explicitly pass" I meant passing them without the eta reduce, in other terms:
swapA' :: (Arrow a) => a ((b,c), (b,c)) (c,b) swapA' t = (????) swapFirst >>> swapSecond (???) where swapFirst = first $ arr snd swapSecond = second $ arr fst
where the question marks indicate that I don't know how to tell swapFirst "hey, even though from the outside I'm passing you a tuple *t*, you have to take as input a (t,t)."
Hope this is clearer or it has some sense at all, maybe I'm not getting correctly the way arrows work!
Perhaps arrow notation (Arrows extension) is what you want: swapA' :: (Arrow a) => a ((b, c), (b, c)) (c, b) swapA' = proc ((_, x), (y, _)) -> id -< (x, y) But I really don't understand why you're implementing this as an arrow computation. A simple function would do the trick, and should you really want to use it in an arrow you can just lift it by applying 'arr'. Greets, Ertugrul -- Not to be or to be and (not to be or to be and (not to be or to be and (not to be or to be and ... that is the list monad.