"Variadic enum cases are not supported" swift 5个错误

Question

我在以下代码中遇到 Variadic enum cases are not supported 错误。这在 Swift4 中编译和工作正常，但在 Swift5 中给出编译时错误，Xcode 10.2

enum ModelRule {
        case required(keys: String...) // error: Variadic enum cases are not supported
        case nonEmptyString(keys: String...) // error: Variadic enum cases are not supported
        case emptyString(key: String)
}

虽然消息很清楚，但我想知道他们为什么要删除完美运行的功能？或者我在这里遗漏了什么？

此外，对于上述错误，有没有比以下更好的解决方案？ case required(keys: [String])

Answer 1

可变参数曾经在 swift 4 上起作用，但这不是故意的。只需使用数组

enum ModelRule {
    case required(keys: [String])
    case nonEmptyString(keys: [String])
    case emptyString(key: String)
}

见Swift Release notes

Answer 2

展开评论：

您应该在案例的关联值中有数组。但是为了方便创建带有可变参数的方法。

示例：

enum ModelRule {
    case required(keys: [String])
    case nonEmptyString(keys: [String])
    case emptyString(key: String)

    init(required keys: String...) { // It could be static func, but init here works and looks better
        self = .required(keys: keys)
    }

    init(nonEmptyString keys: String...) {
        self = .nonEmptyString(keys: keys)
    }

    init(emptyString key: String) { // Added this just for my OCD, not actually required
        self = .emptyString(key: key)
    }
}

用法：

let required = ModelRule(required: "a", "b", "c") // .required(keys: ["a", "b", "c"])
let nonEmpty = ModelRule(nonEmptyString: "d", "e", "f") // .nonEmptyString(keys: ["d", "e", "f"])
let empty = ModelRule(emptyString: "g") // .emptyString(key: "g")

Answer 3

您可以使用静态函数来模拟带有可变参数的情况：

enum ModelRule {
    case required(keys: [String])
    case nonEmptyString(keys: [String])
    case emptyString(key: String)

    static func required(keys: String...) -> Self {
        .required(keys: keys)
    }

    static func nonEmptyString(keys: String...) -> Self {
        .nonEmptyString(keys: keys)
    }
}

用法：

ruleChecker.apply(rule: .required(keys: "a", "b", "c"))

"Variadic enum cases are not supported" swift 5个错误

"Variadic enum cases are not supported" swift 5 error

swift

swift5