Malcolm Wallace wrote:
{-# RULES "foldr2/both" forall k z (g::forall c.(a->c->c)->c->c) (h::forall c.(b->c->c)->c->c) . foldr2 f z (build g) (build h) = g (\x _-> h (\y r-> f x y r) z) z #-}
Looks wrong to me. Somehow it doesn't feel right that the first argument to g doesn't use its sencond argument and I think, what you wrote is actually foldr (f (head (build g))) z (build h) or something like that. Wasn't there some consensus that foldr/build cannot deforest both lists that go into zip? Intuitively it feels right: the "loop" that drives the calculation is contained in the build, and it cannot call out to the other generator to just get one element. Udo. -- "The key to performance is elegance, not battalions of special cases." -- Jon Bentley and Doug McIlroy