Objective C - 声明属性的可选字段

Question

是否有任何解决办法可以在 Objective ci 中找不到任何将 Bool 和 float 类型设置为可选的？

 @property (nonatomic, assign) float percentageOfCases;
    @property (nonatomic, assign) float inspectionPercentageOfCases;
    @property (nonatomic, assign) NSString<Optional> *inspectionStatus;
@property (nonatomic, assign) BOOL sendNotification;

Answer 1

SwiftOptional对应于 Objective-C 中的可空性，而在 Objective-C 中，只有指针类型可以为空。因此，您必须将属性更改为指针类型，在本例中这意味着NSNumber *对于float属性以及NSNumber *或者CFBooleanRef对于BOOL属性。

但是，更改类型将使 Swift 将属性导入为 和NSNumber而CFBoolean不是 asFloat和Bool。因此，您可能还想应用NS_REFINED_FOR_SWIFT到每个属性，并使用 Swift来定义类型和extension的属性。它看起来是这样的：FloatBool

// In your Objective-C header:

@interface MyObject : NSObject
@property (nonatomic, assign, nullable) NSNumber *percentageOfCases NS_REFINED_FOR_SWIFT;
@property (nonatomic, assign, nullable) NSNumber *inspectionPercentageOfCases NS_REFINED_FOR_SWIFT;
@property (nonatomic, assign, nullable) NSString *inspectionStatus;
@property (nonatomic, assign, nullable) CFBooleanRef sendNotification NS_REFINED_FOR_SWIFT;
@end

extension MyObject {
    var percentageOfCases: Float? {
        get { __percentageOfCases?.floatValue }
        set { __percentageOfCases = newValue as NSNumber? }
    }

    var inspectionPercentageOfCases: Float? {
        get { __inspectionPercentageOfCases?.floatValue }
        set { __inspectionPercentageOfCases = newValue as NSNumber? }
    }

    var sendNotification: Bool? {
        get { (__sendNotification as NSNumber?)?.boolValue }
        set { __sendNotification = newValue as NSNumber? }
    }
}

请注意，Xcode 不会为双下划线标识符（如__percentageOfCases）提供代码补全。