斯卡拉更低類型綁定'是方法參數的子類型'限制

Question

在斯卡拉，爲什麼設置方法類型參數的較低類型邊界不強制方法爭論的「是超類型」限制？

object TypeBounds extends App { 
    class MotorVehicle 
    class Truck extends MotorVehicle 
    class Car extends MotorVehicle 
    class Saloon extends Car 
    class HatchBackSaloon extends Saloon 

    def lowerTypeBound[C >: Car](c: C): C = c 

    def upperTypeBound[C <: Car](c: C): C = c 

    // Works. HatchBackSaloon is a sub class of Car 
    println(upperTypeBound(new HatchBackSaloon())) 

    // as expected doesn't compile. Truck is not a subclass of Car  
    println(upperTypeBound(new Truck())) 

    // Compiles and runs, but why ? HatchBackSaloon is not a super class of Car. 
    println(lowerTypeBound(new HatchBackSaloon())) 

} 

Answer 1

C在你的例子是物化爲Car，不HatchbackSaloon。

看起來像def lowerTypeBound(c: Car): Car的函數可以接受類型爲HatchbackSaloon的參數，這並不奇怪，對吧？

嘗試這樣：

val result: HatchBackSaloon = lowerTypeBound(new HatchBackSaloon) 

這不會編譯，因爲它需要C是HatchbackSaloon，這不是一個Car超。但是，這將工作：

val result: MotorVehicle = lowerTypeBound(new HatchbackSaloon) 

因爲C是MotorVehicle這裏，這是允許的。