swift - 如何在 BitwiseOperationsType 中实现运算符

Question

开源代码 LogFlag 类(CocoaLumberjack.swift 不再编译，因为在 Xcode 6 Beta 7 RawOptionSetType 已更改为实现 BitwiseOperationsType。我不知道如何实现运算符功能。

下面的示例使用 CocoaLumberjack.swift 中的 LogFlag:

// RawOptionSetType implements BitwiseOperationsType, so LogFlag won't compile until it implements the operators there

struct LogFlag : RawOptionSetType, BooleanType { 
    private var value: Int32 = 0
    init(_ value: Int32) { self.value = value }
    var boolValue: Bool { return self.value != 0 }
    func toRaw() -> Int32 { return self.value }
    static func fromRaw(raw: Int32) -> LogFlag? { return self(raw) }
    static func fromMask(raw: Int32) -> LogFlag { return self(raw) }
    static func convertFromNilLiteral() -> LogFlag { return self(0) }

    static var Error:   LogFlag { return self(1 << 0) }
    static var Warn:    LogFlag { return self(1 << 1) }
    static var Info:    LogFlag { return self(1 << 2) }
    static var Debug:   LogFlag { return self(1 << 3) }
    static var Verbose: LogFlag { return self(1 << 4) }
}

尝试:

func &(_: LogFlag, _:LogFlag) -> LogFlag { // What goes here? }

这是协议(protocol):

protocol BitwiseOperationsType {
    func &(_: Self, _: Self) -> Self
    func |(_: Self, _: Self) -> Self
    func ^(_: Self, _: Self) -> Self
    prefix func ~(_: Self) -> Self

    /// The identity value for "|" and "^", and the fixed point for "&".
    ///
    /// ::
    ///
    ///   x | allZeros == x
    ///   x ^ allZeros == x
    ///   x & allZeros == allZeros
    ///   x & ~allZeros == x
    ///
    class var allZeros: Self { get }
}

Answer 1

添加就足够了

static var allZeros: LogFlag { return nil }

到结构定义。其余运算符 &、|、^ 和 ~协议(protocol)已定义为通用函数，例如

func &<T : _RawOptionSetType>(a: T, b: T) -> T

如果您确实想要覆盖该运算符，请实现(作为示例)

func &(a: LogFlag, b: LogFlag) -> LogFlag {
    return LogFlag(a.value & b.value)
}