on Tue May 10 2016, Nate Cook <[email protected]> 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.
Why shouldn't this work with all Collections, with an optimized version
for BidirectionalCollections?
> 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
I think we need to consider consistency of naming more carefully in this
area. If we go this route, I want:
x.firstIndex(of: 10)
--
-Dave
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