如何通过在Swift中忽略其关联值来比较枚举与关联值？

问题描述：

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
    }
}

以下代码作品：

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，这不编译：

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

...并提供以下错误消息：

... and it gives the following error message:

二进制运算符'=='不能应用于o CardRank和（Int） - > CardRank'的类型

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

我假设这是因为它期待一个完整的类型而$ code> CardRank.Number 没有指定整个类型，而$ code> CardRank.Number（2）没有。但是，在这种情况下，我希望它匹配任何号码;不仅仅是一个特定的一个。

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，但是实现 == 运算符是避免这种详细的解决方案：

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
}

有没有办法将枚举与关联值进行比较，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 == 方法中的情况更改为 case（.Number，.Number）：return true 虽然它会正确返回，但我的比较仍将看起来像是与特定数字相比较（ number == CardRank.Number（2）;其中2是一个虚拟值）而不是任何号码（ 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).

答

编辑：正如Etan指出的那样，你可以省略（_）通配符匹配，以更干净地使用。

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

不幸的是，我不相信在Swift 1.2中有一个比开关方法更容易的方法。

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

在Swift 2但是，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，可以考虑添加 isNumber 计算属性到你的枚举，实现你的switch语句。

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

如何通过在Swift中忽略其关联值来比较枚举与关联值？

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