Tail-recursive replicateA
#4234
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johnynek
reviewed
Jun 13, 2022
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| map(loop(n))(_.toList) | ||
| map(loop(one, n, pure(Chain.empty)))(_.toList) |
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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
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What if we wrote it as loop(one, n - 1, one) instead?
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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.
| loop( | ||
| productR(fa)(fa), | ||
| n >> 1, | ||
| if ((n & 1) == 1) productR(acc)(fa) else acc |
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same comment here about leveraging productR(unit)(a) == a to avoid doing a needless productR(fa)(unit)
johnynek
approved these changes
Jun 13, 2022
danicheg
approved these changes
Jun 13, 2022
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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.