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Fix keys of a filtered indexed Seq iterated in reverse - #2309
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bdurrer merged 2 commits intoNov 2, 2026
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A filtered indexed Seq renumbers its keys from 0, but when it was iterated backwards the counter still started at 0, so the last element got key 0. lastIndexOf, findLastIndex, findLastKey, findLastEntry and reduceRight then returned positions counted from the end, e.g. Seq([1, 2, 3, 1, 5]).filter((x) => x !== 2).lastIndexOf(1) was 1 instead of 2. Count down from the filtered size instead, materialized with ensureSize like reverseFactory does. Co-Authored-By: Claude Opus 5.5 <[email protected]>
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This sounds a bug indeed, Thanks for your tests and MRs around this! |
bdurrer
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Oct 30, 2026
Co-Authored-By: Claude Opus 5.5 <[email protected]>
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Hm, it fails at least one test |
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That's the |
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A filtered indexed
Seqrenumbers its keys from 0, but when it was iterated backwards the counter still started at 0, so the last element got key 0 and every "last" helper returned positions counted from the end:Listis fine because itsfilteris eager. This is the same symptom #2268 fixed forreverse(): here the filter counts down from its size, materialized withensureSizethe same way. The regression test covers both the__iteratepath (reduceRight,findLast*) and the__iteratorpath (toKeyedSeq().reverse().entries()).I used Claude Code to write this change.