2

我在 Array 上定义一个扩展来覆盖 Slice 创建:

struct S<T> {
    private var array: [T] = []
    private var first = 0
    private var len = 0

    init(_ array: [T], _ range: Range<Int>? = nil) {
        self.array = array
        if let range = range {
            self.first = range.startIndex
            self.len = range.endIndex
        } else {
            self.first = 0
            self.len = array.count
        }
    }
}

extension Array {
    subscript(subRange: Range<Int>) -> S<T> {
        return S<T>(self, subRange)
    }
}

let a = [4, 3, 2, 1, 0, -1][2..<4 as Range<Int>]

但是,我在定义 a 时遇到错误:“范围不可转换为 Int”(没有强制转换,错误是“HalfOpenInterval ...”)。我究竟做错了什么?

4

1 回答 1

1

因为Array已经有了子切片功能:

typealias SubSlice = Slice<T>
subscript (subRange: Range<Int>) -> Slice<T>

因此,为了您的实现工作,您必须明确指定返回类型:

let a = [4, 3, 2, 1, 0, -1][2..<4] as S<Int>
于 2014-11-17T17:03:50.823 回答