二元運算符“+”不能應用於“字符串”和“字符串？”類型的操作數

Question

編程新手，不太了解這是怎么回事，任何建議都值得贊賞。

func exercise() {
    let alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]

    var char0 = alphabet.randomElement()
    var char1 = alphabet.randomElement()
    var char2 = alphabet.randomElement()
    var char3 = alphabet.randomElement()
    var char4 = alphabet.randomElement()
    var char5 = alphabet.randomElement()

    print(char0 + char1 + char2 + char3 + char4 + char5)
}

Answer 1

要理解的錯誤消息的重要部分是 String 和 String? （可選）被編譯器認為是 2 種不同的類型。

在某些情況下，編譯器可以進行（隱式）轉換，因此類型相同但此處不一樣，因為它無法轉換 nil 並且 + 運算符僅適用於相同類型的兩個變量/文字值。

考慮這種情況

var char0 = alphabet.randomElement()
var char1 = alphabet.randomElement() ?? ""

如果我們現在做print("0: \(char0) 1: \(char1)")兩個變量的結果是完全不同的

0：可選（“p”）1：d

如果我們使用print("0: \(type(of: char0)) 1: \(type(of: char1))")檢查他們的類型，我們也會看到

0：可選<字符串> 1：字符串

Answer 2

其他人已經解釋了為什么您的代碼會出錯。 （ String.randomElement()返回一個Optional<String> ，這與 String 不同，您不能使用 + 連接 Optional。）

另一點：您的代碼冗長且重復。 計算機科學中有一個原則“DRY”。 它代表“不要重復自己”。 任何時候，如果您一遍又一遍地使用相同的代碼（可能會有細微的變化），這就是“代碼異味”，也是改進的機會。

你可以像這樣重寫你的代碼而不重復：

func exercise() {
    let alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]

   var result = ""
   for _ in 1...6 {
     result.append(alphabet.randomElement() ?? "")
   }
   print(result)
}

在那個版本中，表達式alphabet.randomElement()只出現一次。 你不要重復自己。 它還使用?? “nil 合並運算符”將可能的 nil 結果轉換為空白字符串。

---

處理它的另一種方法是定義+=運算符的覆蓋，它允許您將 append 字符串選項轉換為字符串：

public func +=(left: inout String, right: Optional<String>) {
    left = left + (right ?? "")
}

然后你的 function 變成：

func exercise() {
    let alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
   var result = ""
   for _ in 1...6 {
    result += alphabet.randomElement()
   }
   print(result)
}

---

另一種方法是打亂您的源數組，獲取前 6 個元素，然后將它們重新組合成一個字符串：

func exercise() {
    let alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
    
    let result = alphabet
        .shuffled()[1...6]
        .joined()
    print(result)
}

Answer 3

通過使用為變量賦予默認值？？ 運算符在 nil 的情況下。 它會改變String的變量類型嗎？ 到字符串

func exercise() {

    let alphabet = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]

    var char0 = alphabet.randomElement() ?? ""
    var char1 = alphabet.randomElement() ?? ""
    var char2 = alphabet.randomElement() ?? ""
    var char3 = alphabet.randomElement() ?? ""
    var char4 = alphabet.randomElement() ?? ""
    var char5 = alphabet.randomElement() ?? ""

    print(char0 + char1 + char2 + char3 + char4 + char5)
}