Tail-recursive replicateA

[armanbilge]

Yet another follow-up :) see #4233 (comment).

It doesn't seem more complex to me than the recursive implementation since it's the same basic algorithm, but I don't feel that strongly. Maybe a benchmark could make it more interesting.

[johnynek]

I have a minor anxiety about the extra cost of the map2(a, pure(Chain.empty))(_.concat(_)) being the same as a.

What if we cheat:

      val emptyF = pure(Chain.empty)

      def concat(left: F[Chain[A]], right: F[Chain[A]]): F[Chain[A]] =
        if (right eq emptyF) left else map2(left, right)(_.concat(_))

      @tailrec def loop(fa: F[Chain[A]], n: Int, acc: F[Chain[A]]): F[Chain[A]] =
         if (n == 1) concat(fa, acc)
         else
           // n >= 2
           // so (n >> 1) >= 1 and we are allowed to call loop
           loop(
             map2(fa, fa)(_.concat(_)),
             n >> 1,
             if ((n & 1) == 1) concat(fa, acc) else acc
           )

So we can leverage the law that concat(a, pure(Chain.empty)) == a

[armanbilge]

What if we wrote it as loop(one, n - 1, one) instead?

[johnynek]

yeah, I think that works... but for replicateA_ I think we need to rewrite so that we only entire the loop in the n >= 2 case as we do here.

[johnynek]

same comment here about leveraging productR(unit)(a) == a to avoid doing a needless productR(fa)(unit)