如何从ClosedRange.SubSequence中删除Slice<>？

Question

我在用chunked method in Swift Algorithms。

当我处理范围时，很容易处理结果类型：

let range: Range<Int> = 0..<5
let subRangesIntermediary: [Range<Int>.SubSequence] = range.chunked(on: { $0 / 2 })
assert(Range<Int>.SubSequence.self == Range<Int>.self)

let subRanges: [Range<Int>] = subRangesIntermediary
assert(subRanges == [0..<2, 2..<4, 4..<5])

这是因为Range<Int>.SubSequence == Range<Int>在两者之间的转换是毫不费力的。

当我尝试用ClosedRange做同样的事情时，我遇到了问题：

let closedRange: ClosedRange<Int> = 0...4
let subClosedRangesIntermediary: [ClosedRange<Int>.SubSequence] = closedRange.chunked(on: { $0 / 2})
let subClosedRangesIntermediary2: [Slice<ClosedRange<Int>>] = subClosedRangesIntermediary
// assert(subClosedRangesIntermediary2 == [0...1, 2...3, 4...4]) // ❌ Binary operator '==' cannot be applied to operands of type '[Slice<ClosedRange<Int>>]' and 'ArraySlice<ClosedRange<Int>>'

那是因为ClosedRange<Int>.SubSequence == Slice<ClosedRange<Int>>。

我希望删除切片，这样就只剩下一个ClosedRange数组。

我想出了一种手动完成的方法，但这看起来需要很多工作：

let subClosedRanges: [ClosedRange<Int>] = subClosedRangesIntermediary2.map {
  let start = closedRange[$0.startIndex]
  let end = closedRange[$0.index(before: $0.endIndex)]
  return start...end
}
assert(subClosedRanges == [0...1, 2...3, 4...4])

我希望有一种方法可以做下面这样的事情：

let subClosedRanges: [ClosedRange<Int>] = subClosedRangesIntermediary2.map { $0.valueInBase }

或者：

// Similar to how we can use Array(slice) on ArraySlice:
let subClosedRanges: [ClosedRange<Int>] = subClosedRangesIntermediary2.map { ClosedRange($0) }

我在Slice上找不到这样的东西。

如何将Slice<ClosedRange<T>>转换为ClosedRange<T>

Answer 1

除非有更权威的答案，否则这里有一个解决方法：

extension ClosedRange where Bound: Strideable, Bound.Stride: SignedInteger {
  init(_ slice: Slice<Self>) {
    let lower = slice.base[slice.startIndex]
    let upper = slice.base[slice.index(before: slice.endIndex)]
    self.init(uncheckedBounds: (lower: lower, upper: upper))
  }
}

用法：

let subClosedRanges: [ClosedRange<Int>] = subClosedRangesIntermediary2.map { ClosedRange($0) }