如何通过忽略 Swift 中的关联值来比较枚举与关联值? [英] How to compare enum with associated values by ignoring its associated value in Swift?

问题描述

阅读如何测试Swift枚举的相等性使用关联值，我实现了以下枚举:

After reading How to test equality of Swift enums with associated values, I implemented the following enum:

enum CardRank {
    case Number(Int)
    case Jack
    case Queen
    case King
    case Ace
}

func ==(a: CardRank, b: CardRank) -> Bool {
    switch (a, b) {
    case (.Number(let a), .Number(let b))   where a == b: return true
    case (.Jack, .Jack): return true
    case (.Queen, .Queen): return true
    case (.King, .King): return true
    case (.Ace, .Ace): return true
    default: return false
    }
}

以下代码有效:

let card: CardRank = CardRank.Jack
if card == CardRank.Jack {
    print("You played a jack!")
} else if card == CardRank.Number(2) {
    print("A two cannot be played at this time.")
}

但是，这不会编译:

let number = CardRank.Number(5)
if number == CardRank.Number {
    print("You must play a face card!")
}

... 并给出以下错误消息:

... and it gives the following error message:

二元运算符'=='不能应用于'CardRank'和'(Int) -> CardRank'类型的操作数

Binary operator '==' cannot be applied to operands of type 'CardRank' and '(Int) -> CardRank'

我假设这是因为它需要一个完整的类型并且 CardRank.Number 没有指定一个完整的类型，而 CardRank.Number(2) 做了.但是，在这种情况下，我希望它匹配任何数字；不只是一个特定的.

I'm assuming this is because it's expecting a full type and CardRank.Number does not specify an entire type, whereas CardRank.Number(2) did. However, in this case, I want it to match any number; not just a specific one.

显然我可以使用 switch 语句，但实现 == 运算符的全部目的是避免这种冗长的解决方案:

Obviously I can use a switch statement, but the whole point of implementing the == operator was to avoid this verbose solution:

switch number {
case .Number:
    print("You must play a face card!")
default:
    break
}

有什么方法可以将枚举与关联值进行比较而忽略其关联值?

Is there any way to compare an enum with associated values while ignoring its associated value?

注意:我意识到我可以将 == 方法中的 case 更改为 case (.Number, .Number): return true，但是，虽然它会正确返回 true，但我的比较看起来仍然像是在与特定数字进行比较(number == CardRank.Number(2); 其中 2 是一个虚拟值)而不是比任何数字(number == CardRank.Number).

Note: I realize that I could change the case in the == method to case (.Number, .Number): return true, but, although it would return true correctly, my comparison would still look like its being compared to a specific number (number == CardRank.Number(2); where 2 is a dummy value) rather than any number (number == CardRank.Number).

推荐答案

正如 Etan 指出的，你可以省略 (_) 通配符匹配来使用这个更干净.

As Etan points out, you can omit the (_) wildcard match to use this more cleanly.

不幸的是，我认为没有比 Swift 1.2 中的 switch 方法更简单的方法了.

Unfortunately, I don't believe that there's an easier way than your switch approach in Swift 1.2.

然而，在 Swift 2 中，您可以使用新的 if-case 模式匹配:

In Swift 2, however, you can use the new if-case pattern match:

let number = CardRank.Number(5)
if case .Number(_) = number {
    // Is a number
} else {
    // Something else
}

如果您想避免冗长，您可以考虑将 isNumber 计算属性添加到实现 switch 语句的枚举中.

If you're looking to avoid verbosity, you might consider adding an isNumber computed property to your enum that implements your switch statement.

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