如何在迅捷中比较Enum?

时间:2014-06-20 10:13:32

标签: ios swift

在Objective-C中

这很好用

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无法在Swift中编译它

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或者

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IOS SDK中的ALAuthorizationStatus定义

enum ALAuthorizationStatus : Int {
    case NotDetermined // User has not yet made a choice with regards to this application
    case Restricted // This application is not authorized to access photo data.
    // The user cannot change this application’s status, possibly due to active restrictions
    //  such as parental controls being in place.
    case Denied // User has explicitly denied this application access to photos data.
    case Authorized // User has authorized this application to access photos data.
}

1 个答案:

答案 0 :(得分:3)

比较运算符==返回Bool,而不是Boolean。 以下编译:

func isAuthorized() -> Bool {
    let status = ALAssetsLibrary.authorizationStatus()
    return status == ALAuthorizationStatus.Authorized
}

(就个人而言,我发现来自Swift编译器的错误消息有时令人困惑。 在这种情况下,问题不是==的参数,而是错误的返回类型。)


实际上,由于自动类型推断,还应编译以下内容:

func isAuthorized() -> Bool {
    let status = ALAssetsLibrary.authorizationStatus()
    return status == .Authorized
}

但它失败了编译错误“找不到会员'授权'”,除非你 明确指定status变量的类型:

func isAuthorized() -> Bool {
    let status:ALAuthorizationStatus = ALAssetsLibrary.authorizationStatus()
    return status == .Authorized
}

这可能是当前Swift编译器中的一个错误(使用Xcode 6 beta 1进行测试)。

更新:第一个版本现在在Xcode 6.1中编译。