ios - 在 Objective-C 中访问 Swift OptionSetType

Question

在我的 Swift 类中，我为实现选项定义了一个 OptionSetType。

struct FulfillmentOption : OptionSetType {
    let rawValue: Int
    
    static let Pickup = FulfillmentOption(rawValue: 1 << 0)
    static let Shipping = FulfillmentOption(rawValue: 1 << 1)
    static let UserShipping = FulfillmentOption(rawValue: 1 << 2)
}

然后我创建一个变量来添加/删除和读取选项。这按预期工作。

 var options: FulfillmentOption = []
 options.insert(FulfillmentOption.Pickup)
 options.contains(FulfillmentOption.Pickup)

但是我需要从我的一个 Objective-C 类访问 options 变量。由于 OptionSetType 未在 Objective-C 中定义，因此该变量对我的任何 Objective-C 类都不可见。

将其公开给 Objective-C 的最佳方式是什么？我应该完全停止使用 OptionSetType 吗？

我考虑过像这样创建 public 和 private 变量以在两者之间进行转换。我不喜欢这个，但这是迄今为止我想出的最好的。

private var _options: FulfillmentOptions = []
private var options: UInt {
  get {
    // get raw value from _options
  }
  set {
    // set raw value to _options
  }
}

有没有更优雅的方法来完成这个？我想避免编写不必要的代码。

Answer 1

不是您问题的直接答案，但您可以作为替代方案反过来工作。定义

typedef NS_OPTIONS(NSInteger, FulfillmentOption) {
    FulfillmentOptionPickup = 1 << 0,
    FulfillmentOptionShipping = 1 << 1,
    FulfillmentOptionUserShipping = 1 << 2,
};

在 Objective-C header 中，这将被导入到 Swift 中作为

public struct FulfillmentOption : OptionSetType {
    public init(rawValue: Int)

    public static var Pickup: FulfillmentOption { get }
    public static var Shipping: FulfillmentOption { get }
    public static var UserShipping: FulfillmentOption { get }
}

更多信息可以在 "Using Swift with Cocoa and Objective-C" 中找到引用:

• “与 C API 交互”:

Swift also imports C-style enumerations marked with the NS_OPTIONS macro as a Swift option set. Option sets behave similarly to imported enumerations by truncating their prefixes to option value names.

• “同一项目中的 Swift 和 Objective-C”:

You’ll have access to anything within a class or protocol that’s marked with the @objc attribute as long as it’s compatible with Objective-C. This excludes Swift-only features such as those listed here:

• ...
• Structures defined in Swift
• ...