swift – 子类是否可以覆盖函数并进行更严格的返回？

class Pet {}

class Dog: Pet {}

class House {
    func getPets() -> Pet {
        return Pet()
    }
}

class DogHouse: House {
    override func getPets() -> Dog {
        return Dog()
    }
}

然而,这里的问题是Array< Dog>与Array< Pet>不是“更受限制”的类型,相反,Array< Pet>不是Array< Dog>的推广.形式上,Array< Dog>不是Array< Pet>的协变量.

为了说明原因,请考虑以下示例：

class House<T> {
    var occupants = [T]()

    func addOccupant(_ o: T) {
        occupants.append(o)
    }
}

class Pet {}
class Dog: Pet {}
class Cat: Pet {}

class PetHouseBuilder {
    func buildHouse() -> House<Pet> {
        return House()
    }
}

class DogHouseBuilder: PetHouseBuilder {
    // Suppose this were legal
    override func buildHouse() -> House<Dog> {
        return House()
    }
}

// The concrete return type of the object is `House<Dog>`,but
// `PetHouseBuilder.buildHouse` has a static return type of `House<Pet>`,// so `petHouse` will have an inferred static type of `House<Pet>`
let petHouse = PetHouseBuilder().buildHouse()

let vulnerableLittle