fix 2.19 Nested

source/chapter2/19_Nested_Types.md

@@ -1,14 +1,13 @@
-> 翻译：Lin-H
-
-> 校对：shinyzhu
+> 翻译：Lin-H  
+> 校对：shinyzhu  
 
 # 类型嵌套
 -----------------
 
 本页包含内容：
 
--   [类型嵌套实例](#nested_types_in_action)
--   [类型嵌套的引用](#referring_to_nested_types)
+- [类型嵌套实例](#nested_types_in_action)
+- [类型嵌套的引用](#referring_to_nested_types)
 
 枚举类型常被用于实现特定类或结构体的功能。也能够在有多种变量类型的环境中，方便地定义通用类或结构体来使用，为了实现这种功能，Swift允许你定义类型嵌套，可以在枚举类型、类和结构体中定义支持嵌套的类型。
 
@@ -21,41 +20,45 @@
 
 在`BlackjackCard`规则中，`Ace`牌可以表示1或者11，`Ace`牌的这一特征用一个嵌套在枚举型`Rank`的结构体`Values`来表示。
 
-    struct BlackjackCard {
-        // 嵌套定义枚举型Suit
-        enum Suit: Character {
-           case Spades = "♠", Hearts = "♡", Diamonds = "♢", Clubs = "♣"
-       }
-        // 嵌套定义枚举型Rank
-        enum Rank: Int {
-           case Two = 2, Three, Four, Five, Six, Seven, Eight, Nine, Ten
-           case Jack, Queen, King, Ace
-           struct Values {
-               let first: Int, second: Int?
-           }
-           var values: Values {
-            switch self {
-            case .Ace:
-                return Values(first: 1, second: 11)
-            case .Jack, .Queen, .King:
-                return Values(first: 10, second: nil)
-            default:
-                return Values(first: self.toRaw(), second: nil)
-                }
-           }
-        }
-        // BlackjackCard 的属性和方法
-        let rank: Rank, suit: Suit
-        var description: String {
-        var output = "suit is \(suit.toRaw()),"
-            output += " value is \(rank.values.first)"
-            if let second = rank.values.second {
-                output += " or \(second)"
-            }
-            return output
-        }
+```swift
+struct BlackjackCard {
+    // 嵌套定义枚举型Suit
+    enum Suit: Character {
+       case Spades = "♠", Hearts = "♡", Diamonds = "♢", Clubs = "♣"
     }
 
+    // 嵌套定义枚举型Rank
+    enum Rank: Int {
+       case Two = 2, Three, Four, Five, Six, Seven, Eight, Nine, Ten
+       case Jack, Queen, King, Ace
+       struct Values {
+           let first: Int, second: Int?
+       }
+       var values: Values {
+        switch self {
+        case .Ace:
+            return Values(first: 1, second: 11)
+        case .Jack, .Queen, .King:
+            return Values(first: 10, second: nil)
+        default:
+            return Values(first: self.toRaw(), second: nil)
+            }
+       }
+    }
+    
+    // BlackjackCard 的属性和方法
+    let rank: Rank, suit: Suit
+    var description: String {
+    var output = "suit is \(suit.toRaw()),"
+        output += " value is \(rank.values.first)"
+        if let second = rank.values.second {
+            output += " or \(second)"
+        }
+        return output
+    }
+}
+```
+
 枚举型的`Suit`用来描述扑克牌的四种花色，并分别用一个`Character`类型的值代表花色符号。
 
 枚举型的`Rank`用来描述扑克牌从`Ace`~10,`J`,`Q`,`K`,13张牌，并分别用一个`Int`类型的值表示牌的面值。(这个`Int`类型的值不适用于`Ace`,`J`,`Q`,`K`的牌)。
@@ -71,9 +74,11 @@
 
 因为`BlackjackCard`是一个没有自定义构造函数的结构体，在[Memberwise Initializers for Structure Types](https://github.com/CocoaChina-editors/Welcome-to-Swift/blob/master/The%20Swift%20Programming%20Language/02Language%20Guide/14Initialization.md)中知道结构体有默认的成员构造函数，所以你可以用默认的`initializer`去初始化新的常量`theAceOfSpades`:
 
-    let theAceOfSpades = BlackjackCard(rank: .Ace, suit: .Spades)
-    println("theAceOfSpades: \(theAceOfSpades.description)")
-    // 打印出 "theAceOfSpades: suit is ♠, value is 1 or 11"
+```swift
+let theAceOfSpades = BlackjackCard(rank: .Ace, suit: .Spades)
+println("theAceOfSpades: \(theAceOfSpades.description)")
+// 打印出 "theAceOfSpades: suit is ♠, value is 1 or 11"
+```
 
 尽管`Rank`和`Suit`嵌套在`BlackjackCard`中，但仍可被引用，所以在初始化实例时能够通过枚举类型中的成员名称单独引用。在上面的例子中`description`属性能正确得输出对`Ace`牌有1和11两个值。
 
@@ -82,7 +87,10 @@
 
 在外部对嵌套类型的引用，以被嵌套类型的名字为前缀，加上所要引用的属性名：
 
-    let heartsSymbol = BlackjackCard.Suit.Hearts.toRaw()
-    // 红心的符号 为 "♡"
+```swift
+let heartsSymbol = BlackjackCard.Suit.Hearts.toRaw()
+// 红心的符号 为 "♡"
+```
 
 对于上面这个例子，这样可以使`Suit`, `Rank`, 和 `Values`的名字尽可能的短，因为它们的名字会自然的由被定义的上下文来限定。
+