String不实现RandomAccessIndexType。可能是因为它们支持具有不同字节长度的字符。这就是为什么我们必须使用string.characters.count(在Swift 1.x中的count或countElements)来获取字符数。这也适用于职位。 _position可能是一个索引到原始数组的字节，他们不想暴露。 String.Index旨在保护我们在字符中间访问字节。

这意味着您获得的任何索引必须从String.startIndex或String.endIndex(String.Index实现BidirectionalIndexType)创建。任何其他索引可以使用后继或前导方法创建。

现在为了帮助我们索引，有一组方法(Swift 1.x中的函数)：

Swift 3.0

let text = "abc"
let index2 = text.index(text.startIndex,offsetBy: 2) //will call succ 2 times
let lastChar: Character = text[index2] //now we can index!

let characterIndex2 = text.characters.index(text.characters.startIndex,offsetBy: 2)
let lastChar2 = text.characters[characterIndex2] //will do the same as above

let range: Range<String.Index> = text.range(of: "b")!
let index: Int = text.distance(from: text.startIndex,to: range.lowerBound)

Swift 2.x

let text = "abc"
let index2 = text.startIndex.advancedBy(2) //will call succ 2 times
let lastChar: Character = text[index2] //now we can index!
let lastChar2 = text.characters[index2] //will do the same as above

let range: Range<String.Index> = text.rangeOfString("b")!
let index: Int = text.startIndex.distanceTo(range.startIndex) //will call successor/predecessor several times until the indices match

Swift 1.x

let text = "abc"
let index2 = advance(text.startIndex,2) //will call succ 2 times
let lastChar: Character = text[index2] //now we can index!

let range = text.rangeOfString("b")
let index: Int = distance(text.startIndex,range.startIndex) //will call succ/pred several times

使用String.Index是麻烦的，但使用包装器以整数索引(见http://stackoverflow.com/a/25152652/669586)是危险的，因为它隐藏了真实索引的低效率。

请注意，Swift索引实现存在的问题是，为一个字符串创建的索引/范围不能可靠地用于不同的字符串，例如：

Swift 2.x

let text: String = "abc"
let text2: String = "???"

let range = text.rangeOfString("b")!

//can randomly return a bad substring or throw an exception
let substring: String = text2[range]

//the correct solution
let intIndex: Int = text.startIndex.distanceTo(range.startIndex)
let startIndex2 = text2.startIndex.advancedBy(intIndex)
let range2 = startIndex2...startIndex2

let substring: String = text2[range2]

Swift 1.x

let text: String = "abc"
let text2: String = "???"

let range = text.rangeOfString("b")

//can randomly return nil or a bad substring 
let substring: String = text2[range] 

//the correct solution
let intIndex: Int = distance(text.startIndex,range.startIndex)    
let startIndex2 = advance(text2.startIndex,intIndex)
let range2 = startIndex2...startIndex2

let substring: String = text2[range2]