1 //可以用来快速的枚举数组中的内容，并用元组(index, value)接收
2 public func enumerated() -> EnumeratedSequence<Array<Element>>
3 
4 //可以倒叙数组内容
5 public func reversed() -> ReversedRandomAccessCollection<Array<Element>>

 1 let arrayStr:[String] = ["s","t","r","e","n","g","t","h","e","n","!"]
 2 
 3 //索引元素值绑定一起倒序遍历
 4 for i in arrayStr.enumerated().reversed() {
 5   print("\(i) ", terminator: "")
 6 }
 7 /*
 8 (offset: 10, element: "!") (offset: 9, element: "n") (offset: 8, element: "e") (offset: 7, element: "h") (offset: 6, element: "t") (offset: 5, element: "g") (offset: 4, element: "n") (offset: 3, element: "e") (offset: 2, element: "r") (offset: 1, element: "t") (offset: 0, element: "s") 
 9 */
10 
11 //只元素值倒序遍历，原索引不变
12 for i in arrayStr.reversed().enumerated() {
13   print("\(i) ", terminator: "")
14 }
15 /*
16 (offset: 0, element: "!") (offset: 1, element: "n") (offset: 2, element: "e") (offset: 3, element: "h") (offset: 4, element: "t") (offset: 5, element: "g") (offset: 6, element: "n") (offset: 7, element: "e") (offset: 8, element: "r") (offset: 9, element: "t") (offset: 10, element: "s") 
17 */

 1 let arrayStr:[String] = ["s","t","r","e","n","g","t","h","e","n","!"]
 2 
 3 //方法1:根据元素倒序遍历
 4 for item in arrayStr.reversed() {
 5     print(item)
 6 }
 7 
 8 //方法2:根据部分元素倒序遍历
 9 for item in arrayStr[1...5].reversed() {
10     print(item)
11 }
12 
13 //方法3：根据索引倒序遍历
14 for index in (0...(arrayStr.count - 1)).reversed() {
15     print(arrayStr[index])
16 }
17 
18 //方法4:初始化数组并倒序遍历
19 for item in Array(1...10).reversed() {
20     print(item)
21 }
22 
23 //方法4:初始化数组并倒序遍历
24 for item in Array(1...10).reversed() {
25     print(item)
26 }
27 
28 //方法5：stride函数
29 //from，to，by 最后一个值是 >(或<) to的值
30 //from，through，by 最后一个值是 >=(或<=) through的值
31 for index in stride(from: arrayStr.count - 1, through: 0,by: -1 )
32 {
33      print(arrayStr[index])
34 }

1 let arrayStr:Set<String> = ["s","t","r","e","n","g"]
2 //按照字母排序倒序输出
3 for item in arrayStr.sorted().reversed() {
4     print("\(item) ", terminator: "")
5 }
6 //Print "t s r n g e" 

1 var direc: [Int: String] =  [0:"s",1:"t",2:"r",3:"e",4:"n",5:"g"]
2 //按键值对倒序遍历
3 for (num, char) in direc.sorted(by: <).reversed() {
4     print("\(num): \(char)")
5 }

我发布了几个用 Swift 2.0和2.1编写的应用程序.现在Apple刚开源Swift并宣布Swift 3.0将于明年发布.在Swift 3.0发布后,我的应用程序仍然有用吗？

我想你在这里问两个不同的问题.

>当Swift 3推出时,您的应用程序是否会在新操作系统上执行？
>推出时,您的应用会在新操作系统上运行吗？

关于第一个问题,是的,你的应用程序将执行,即使它是针对Swift 2编译的.Swift应用程序在其捆绑包中嵌入了Swift运行时库,因此您在这方面是安全的.您会注意到Swift捆绑包总是比Objective-C对应物大4-5 MB,这正是原因所在.

关于第二个问题,与任何新的操作系统版本一样,无法保证您的应用程序能够像以前版本的操作系统那样工作.虽然根据我的经验,Apple在过去的几个OS版本中做了相当不错的工作,尽管你的里程可能会有所不同.