scala – 函数在其参数类型中是逆变的,在返回类型中是共变量

我理解协变和逆变独立意味着什么.

如果我有以下课程：

class Car {}
class SportsCar extends Car {}
class Ferrari extends SportsCar {}

和：

object covar extends App {

    // Test 1: Works as expected

    def test1( arg: SportsCar => SportsCar ) = {
        new SportsCar
    }

    def foo1(arg: Car): Ferrari = { new Ferrari }
    def foo2(arg: SportsCar): Car = { new Ferrari }
    def foo3(arg: Ferrari): Ferrari = { new Ferrari }

    test1(foo1) // compiles
    test1(foo2) // Fails due to wrong return type - violates return type is covariant
    test1(foo3) // Fails due to wrong parameter type - violates param type is contravariant

    // Test 2: Confused - why can I call test2 succesfully with ferrari 
    // as the parameter,and unsuccesfully with a car as the parameter?

    def test2(arg: SportsCar): SportsCar = {
        new Ferrari
    }

    val car = new Car()
    val sportsCar = new SportsCar()
    val ferrari = new Ferrari()

    val f1 = test2(ferrari)  // compiles - why?
    val f2 = test2(car)      // fails - why?

}

如上所述,在测试2中,为什么我可以使用法拉利作为参数成功调用test2,并且以汽车作为参数不成功？

语句Function是否在其参数类型中是逆变的,并且返回类型中的co-variant仅适用于作为参数传递的函数？我想我没有在声明和我写的2个测试之间做出适当的区分.