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Absent semigroup should short-circut folds on first nothing

See original GitHub issue

Absent#foldLeft and #foldRight should process the minimal number of elements of the inbound Iterable necessary to produce the result, and since the semigroup treats nothing as an annihilator, that means that nothing() should cause a short-circuit. As a good heuristic, both Absent.<Unit>absent(Constantly::constantly).foldLeft(nothing(), repeat(just(UNIT))); and Absent.<Unit>absent(Constantly::constantly).foldRight(nothing(), repeat(just(UNIT))).value(); should terminate.

Issue Analytics

State:
Created 3 years ago
Comments:10 (10 by maintainers)

Top GitHub Comments

1reaction

jnapecommented, Mar 13, 2021

@corlaez Some combination of FoldRight.foldRight and lazyZip ought to do it. Something like:

FoldRight.foldRight(
        (maybeX, acc) -> maybeX.lazyZip(acc.fmap(maybeY -> maybeY.fmap(aSemigroup.flip()))),
        lazy(acc),
        rest);

Combine that with an initial match on the accumulator (so knowledge about a nothing() accumulator can be leveraged to elide the entire foldRight) and I think you’re home.

Makes sense?

0reactions

jnapecommented, Mar 15, 2021

Closed by #116