arrays - 闭包数组简写 : Xcode Swift: Expression too complex to be solved in a reasonable time

Question

我正在进行简单的转换(例如英制到公制)。我创建了一个包含 (String, Closure) 元素的元组数组。这让我可以将一个字符串用于转换(例如“英里到公里”到选择器控件中，并在一行中引用关联的公式，传递要转换的值。我遇到的问题是当我使用 $0闭包速记。我收到错误“表达式太复杂，无法在合理的时间内解决。”

这是有效的声明和代码:

在初始类语句下面用 ViewController 属性声明:

var formulaTuple = [(convString: String, convFormula: ((Double) -> Double))]()

在 viewDidLoad() 中赋值

    formulaTuple = [("miles to kilometers", {(a: Double) -> Double in return (a / 0.62137) }),
               ("kilometers to miles", {(a: Double) -> Double in return (a * 0.62137) }),
               ("feet to meters", {(a: Double) -> Double in return (a / 3.2808) }),
               ("yards to meters", {(a: Double) -> Double in return (a / 1.0936) }),
               ("meters to feet", {(a: Double) -> Double in return (a * 3.2808) }),
               ("meters to yards", {(a: Double) -> Double in return (a * 1.0936) }),
               ("inches to centimeters", {(a: Double) -> Double in return (a / 0.39370) }),
               ("centimeters to inches", {(a: Double) -> Double in return (a * 0.39370) }),
               ("fahrenheit to celsius", {(a: Double) -> Double in return ((a - 32) * (5/9)) }),
               ("celsius to fahrenheit", {(a: Double) -> Double in return (a * (9/5) + 32) }),
               ("quarts to liters", {(a: Double) -> Double in return (a / 1.05669) }),
               ("liters to quarts", {(a: Double) -> Double in return (a * 1.05669) }) ]

代码中的工作调用，其中 row 是选择器中被点击的行，inputValue 是传递给要转换的内容，outputValue 是转换的结果。

outputValue = formulaTuple[row].convFormula(inputValue)

当我尝试在 viewDidLoad() 中使用此语法而不是上面的语法来精简声明时出现问题:

    formulaTuple = [("miles to kilometers", {$0 / 0.62137 }),
                    ("kilometers to miles", {$0 * 0.62137 }),
                    ("feet to meters", {$0 / 3.2808 }),
                    ("yards to meters", {$0 / 1.0936 }),
                    ("meters to feet", {$0 * 3.2808 }),
                    ("meters to yards", {$0 * 1.0936 }),
                    ("inches to centimeters", {$0 / 0.39370}),
                    ("centimeters to inches", {$0 * 0.39370 }),
                    ("fahrenheit to celsius", {($0 - 32) * (5/9) }),
                    ("celsius to fahrenheit", {$0 * (9/5) + 32 }),
                    ("quarts to liters", {$0 / 1.05669 }),
                    ("liters to quarts", {$0 * 1.05669 }) ]

我原以为这样会更流畅，但它似乎破坏了 Xcode。想法？我的方法从根本上来说是不合理的，会推荐一种不同的方法吗？谢谢!

Answer 1

当你给 Swift 大数组字面量然后期望它解释类型时，Swift 表现不佳。在你的情况下，我希望它能工作，因为你的 formulaTuple 属性已经有一个确定的类型。

作为解决方法，您可以先初始化一个常量数组，然后将其分配给您的属性:

let temp: [(String, (Double) -> Double)] = [
            ("miles to kilometers", {$0 / 0.62137 }),
            ("kilometers to miles", {$0 * 0.62137 }),
            ("feet to meters", {$0 / 3.2808 }),
            ("yards to meters", {$0 / 1.0936 }),
            ("meters to feet", {$0 * 3.2808 }),
            ("meters to yards", {$0 * 1.0936 }),
            ("inches to centimeters", {$0 / 0.39370}),
            ("centimeters to inches", {$0 * 0.39370 }),
            ("fahrenheit to celsius", {($0 - 32) * (5/9) }),
            ("celsius to fahrenheit", {$0 * (9/5) + 32 }),
            ("quarts to liters", {$0 / 1.05669 }),
            ("liters to quarts", {$0 * 1.05669 })
]

formulaTuple = temp

---

替代答案

元组实际上用于临时存储并从一个函数传回多个结果。您可能需要考虑使用 struct 作为数组值的类型:

struct Conversion {
    let string: String
    let formula: (Double) -> Double
}

var conversions = [Conversion]()

conversions = [
    Conversion(string: "miles to kilometers", formula: {$0 / 0.62137 }),
    Conversion(string: "kilometers to miles", formula: {$0 * 0.62137 }),
    Conversion(string: "feet to meters", formula: {$0 / 3.2808 }),
    Conversion(string: "yards to meters", formula: {$0 / 1.0936 }),
    Conversion(string: "meters to feet", formula: {$0 * 3.2808 }),
    Conversion(string: "meters to yards", formula: {$0 * 1.0936 }),
    Conversion(string: "inches to centimeters", formula: {$0 / 0.39370}),
    Conversion(string: "centimeters to inches", formula: {$0 * 0.39370 }),
    Conversion(string: "fahrenheit to celsius", formula: {($0 - 32) * (5/9) }),
    Conversion(string: "celsius to fahrenheit", formula: {$0 * (9/5) + 32 }),
    Conversion(string: "quarts to liters", formula: {$0 / 1.05669 }),
    Conversion(string: "liters to quarts", formula: {$0 * 1.05669 })
]

outputValue = conversions[row].formula(inputValue)

Swift 对此更满意，temp 解决方法不是必需的。