Clarify endIndex behavior of partitioningIndex(where:) in documentation (Resolves #170)
#171
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mdznr
commented
Oct 23, 2021
| /// | ||
| /// The collection must already be partitioned according to the predicate. | ||
| /// That is, there should be an index `i` where for every element in | ||
| /// `collection[..<i]` the predicate is `false`, and for every element in | ||
| /// `collection[i...]` the predicate is `true`. | ||
| /// | ||
| /// - Parameter belongsInSecondPartition: A predicate that partitions the | ||
| /// collection. | ||
| /// - Returns: The index of the first element in the collection for which |
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I’m still a bit unsure about this sentence. It almost sounds like the first time that predicate is evaluated and returns true, execution stops and that index is returned, like firstIndex(where:).
mdznr
commented
Oct 23, 2021
xwu
reviewed
Oct 23, 2021
natecook1000
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Dec 3, 2021
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natecook1000
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Dec 4, 2021
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@swift-ci Please test |
The returned index isn’t necessarily an index of an element. In the case where it returns `endIndex`, it’s not valid for subscripting the collection. Resolves apple#170
…llection belong in one of the two partitions
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@swift-ci Please test |
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Resolves #170
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