string - 从常量 Ints 创建 Range

Question

这段构建范围的代码有什么问题，然后调用 substringWithRange？

let range = Range<String.Index>(start: 0, end: 3)

Swift 编译器(在 Xcode 7.1.1 中)用此错误消息标记它:

Cannot invoke initializer for type 'Range<Index>' with an argument list of type '(start: Int, end: Int)'

Answer 1

需要引用具体字符串的startIndex，然后前进:

let longString = "Supercalifragilistic"
let startIndex = longString.startIndex
let range = Range(start: startIndex, end: startIndex.advancedBy(3))