Swift 3.0 OptionSet 位测试

Question

使用以下选项集:

struct StatusOptions : OptionSet {
    let rawValue: Int
    static let CountdownDuration0 = StatusOptions(rawValue: 1 << 0)
    static let CountdownDuration1 = StatusOptions(rawValue: 1 << 1)
    static let CountdownDuration2 = StatusOptions(rawValue: 1 << 2)
    static let CountdownDuration3 = StatusOptions(rawValue: 1 << 3)
    static let CountdownDuration4 = StatusOptions(rawValue: 1 << 4)
    static let CountdownDuration5 = StatusOptions(rawValue: 1 << 5)
    static let HomeMode = StatusOptions(rawValue: 1 << 6)
    static let AwayMode = StatusOptions(rawValue: 1 << 7)

    static let Disarmed: StatusOptions = []
    static let ArmedHome: StatusOptions = .HomeMode
    static let ArmedAway: StatusOptions = .AwayMode
    static let ArmedBoth: StatusOptions = [.HomeMode, .AwayMode]
    static let ArmingCountdown: StatusOptions = [.CountdownDuration0, .CountdownDuration1, .CountdownDuration2, .CountdownDuration3, .CountdownDuration4, .CountdownDuration5]
}

如何比较一个字节

var statusFlags: UInt8 = 0b00000000

到选项？

以下是我认为可行的方法，但会导致 playground 错误:

if (statusFlags & .Disarmed) {
    print("Disarmed")
}

谢谢

Answer 1

因为 .Disarmed 是零，你说:

if statusFlags == .Disarmed

或者更简单:

if statusFlags.isEmpty

对于其他值，您说，例如:

if statusFlags.contains(.CountdownDuration3)

这就是 OptionSet 的全部点。您使用集合 操作，而不是算术位操作。

但是要让它起作用，以便您可以利用 OptionSet 的优点，statusFlags 需要 StatusOptions。您可以通过 StatusOptions rawValue: initializer:

来做到这一点

let statusFlags = StatusOptions(rawValue:0b00000000)
statusFlags.isEmpty // true

(请注意，在 Swift 3 中，可能甚至不会使用像 .Disarmed 这样的零位掩码。零只是一个空集。)