无法使用具体的子类在Swift中的协议中实现属性

Pride Chung 发表于 Dev

钟ride

我在做类似于JAVA桥接模式的东西，DriverType是一个协议，要求指定属性vehicle是一个Drivable对象，Drivable也是一个协议，由级“轿车”通过

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'
}

当我实现Driver该类时，将vehicleproperty声明为是很自然的Car。但是后来我遇到了一个问题，即即使完全Driver符合要求，编译器也认为不符合要求。DriverTypeCarDrivable

更新：

@Antonio的答案很可靠，但这是我目前解决的问题，它与类无关。

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
}

安东尼奥

我认为编译器错误会引起误解。DriverType声明采用它的任何类都必须公开vehicle具有Drivable类型的属性，而不是具有采用该Drivable类型的类类型的属性。

我将通过使用泛型定义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() 
}