二元运算符“+”不能应用于“字符串”和“字符串？”类型的操作数

Question

编程新手，不太了解这是怎么回事，任何建议都值得赞赏。

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

要理解的错误消息的重要部分是 String 和 String? （可选）被编译器认为是 2 种不同的类型。

在某些情况下，编译器可以进行（隐式）转换，因此类型相同但此处不一样，因为它无法转换 nil 并且 + 运算符仅适用于相同类型的两个变量/文字值。

考虑这种情况

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

如果我们现在做print("0: \(char0) 1: \(char1)")两个变量的结果是完全不同的

0：可选（“p”）1：d

如果我们使用print("0: \(type(of: char0)) 1: \(type(of: char1))")检查他们的类型，我们也会看到

0：可选<字符串> 1：字符串

Answer 2

其他人已经解释了为什么您的代码会出错。 （ String.randomElement()返回一个Optional<String> ，这与 String 不同，您不能使用 + 连接 Optional。）

另一点：您的代码冗长且重复。 计算机科学中有一个原则“DRY”。 它代表“不要重复自己”。 任何时候，如果您一遍又一遍地使用相同的代码（可能会有细微的变化），这就是“代码异味”，也是改进的机会。

你可以像这样重写你的代码而不重复：

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)
}

在那个版本中，表达式alphabet.randomElement()只出现一次。 你不要重复自己。 它还使用?? “nil 合并运算符”将可能的 nil 结果转换为空白字符串。

---

处理它的另一种方法是定义+=运算符的覆盖，它允许您将 append 字符串选项转换为字符串：

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

然后你的 function 变成：

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)
}

---

另一种方法是打乱您的源数组，获取前 6 个元素，然后将它们重新组合成一个字符串：

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

通过使用为变量赋予默认值？？ 运算符在 nil 的情况下。 它会改变String的变量类型吗？ 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)
}