Question

我在JAVA中执行类似于Bridge Pattern的操作，DriverType是协议，需要名为vehicle的属性为Drivable对象，Drivable也是一个协议，并被班级＆＃39; Car＆＃39;

采用

protocol Drivable {
    var speed: Double { get }
}

protocol DriverType {
    var vehicle: Drivable { get }
}

class Car: Drivable {
    var speed = 80.0;
    var brand = "BMW"
}

class Driver: DriverType {

    var vehicle: Car = Car() //Error: Type 'Driver' does not conform to protocol 'DriverType'

    // This one works, but I have to downcast the property to 'Car' everytime I use it.
    var vehicle: Drivable = Car() //Type 'Driver' does not conform to protocol 'DriverType'
}

当我实施Driver课程时，将vehicle属性声明为Car非常自然。但后来我遇到了编译器认为Driver不符合DriverType的问题，即使Car完全符合Drivable。

更新：

@Antonio的回答很可靠，但这是我目前所解决的问题，它并不涉及到该类的泛型。

protocol Drivable {
    var speed: Double { get }
    init()
}

protocol DriverType {
    func vehicle() -> Drivable
}

class Car: Drivable {
    var speed = 80.0;
    var brand = "BMW"
    required init() {}
}

class Driver: DriverType {
    private var m_vehicle: Car = Car()
    func vehicle() -> Drivable {
        return m_vehicle
    }

    // Inside Driver class I use the m_vehicle property directly
}

Answer 1

我认为编译错误具有误导性。 DriverType声明任何采用它的类都必须使用vehicle类型公开Drivable属性，而不是使用Drivable类型的类类型的属性。

我会通过使用泛型定义DriverType协议和Car类来解决此问题：

protocol Drivable {
    var speed: Double { get }
    init()
}

protocol DriverType {
    typealias T: Drivable
    var vehicle: T { get }
}

class Car: Drivable {
    var speed = 80.0;
    var brand = "BMW"
    required init() {}
}

class Driver<T: Drivable>: DriverType {   
    var vehicle: T = T() 
}

这明确指出采用DriverType的类必须公开vehicle属性，其类型是采用Drivable协议的任何类。

不能使用具体的子类在Swift中实现协议中的属性

1 个答案: