Do these return endIndex or nil on a not-found?
Either can work with different tradeoffs; depends on what you care about...but
the choice should be justified and the alternative explained.
> On May 10, 2016, at 13:54, Nate Cook via swift-evolution
> <swift-evolution@swift.org> wrote:
>
> I've needed these in the past and used them in other languages—any feedback
> on this idea?
>
> Add last(where:) and lastIndex(where:) Methods to Bidirectional Collections
> The standard library should include methods for finding the last element of a
> bidirectional collection that matches a predicate, along with the index of
> that element.
>
> Motivation
> The standard library currently has (or will soon have) methods that perform a
> linear search from the beginning of a collection to find an element that
> matches a predicate:
>
> let a = [20, 30, 10, 40, 20, 30, 10, 40, 20]
> a.first(where: { $0 > 25 }) // 30
> a.index(of: 10) // 2
> a.index(where: { $0 > 25 }) // 1
> Unfortunately, there is no such method that searches from the end of a
> bidirectional collection. Finding the last of particular kind of element has
> multiple applications, particularly with text, such as wrapping a long string
> into lines of a maximum length or trimming whitespace from the beginning and
> end of a string.
>
> This limitation can be worked around by using the methods above on the
> reversed collection, but the resulting code is truly dreadful. For example,
> to find the corresponding last index to a.index(where: { $0 > 25 }), this
> unholy incantation is required:
>
> (a.reversed().index(where: { $0 > 25 })?.base).flatMap({ a.index(before: $0)
> })
> Wat.
>
> Proposed solution
> Bidirectional collections should include three new methods for symmetry with
> the existing forward-searching APIs: last(where:), lastIndex(where:), and
> lastIndex(of:), specifically for collections of Equatable elements.
>
> These additions would remove the need for searching in a reversed collection
> and allow code like the following:
>
> a.last(where: { $0 > 25 }) // 40
> a.lastIndex(of: 10) // 6
> a.lastIndex(where: { $0 > 25 }) // 7
> Much better!
>
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