斯卡拉更低类型绑定'是方法参数的子类型'限制

Question

在斯卡拉，为什么设置方法类型参数的较低类型边界不强制方法争论的“是超类型”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在你的例子是物化为Car，不HatchbackSaloon。

看起来像def lowerTypeBound(c: Car): Car的函数可以接受类型为HatchbackSaloon的参数，这并不奇怪，对吧？

尝试这样：

val result: HatchBackSaloon = lowerTypeBound(new HatchBackSaloon) 

这不会编译，因为它需要C是HatchbackSaloon，这不是一个Car超。但是，这将工作：

val result: MotorVehicle = lowerTypeBound(new HatchbackSaloon) 

因为C是MotorVehicle这里，这是允许的。