从常量 Int 创建 Range<String.Index>

Question

这段构建范围的代码有什么问题，这个范围应该在对 substringWithRange 的调用中提供？

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))

Answer 2

你可以使用多种方式

let startIndex = text.startIndex
let endIndex = text.endIndex

var range1 = startIndex.advancedBy(1) ..< text.endIndex.advancedBy(-4)
var range2 = startIndex.advancedBy(0) ..< startIndex.advancedBy(5)
var range3 = startIndex ..< endIndex

让子串 = text.substringWithRange(范围)

从常量 Int 创建 Range<String.Index>

Creating Range<String.Index> from constant Ints

string

range

swift

swift2