scala协方差,逆方差混淆

我是斯卡拉的新人,真的很困惑。请帮帮我。

/**
2    * Remember! In Scala, every function that takes one argument 
3    * is an instance of Function1 with signature:
4    *
5    * trait Function1[-T, +S] extends AnyRef
6    */
7   
8   class Vehicle(val owner: String)
9   class Car(owner: String) extends Vehicle(owner)
10  
11  object Printer {
12
13    val cars = List(new Car("john"), new Car("paul"))
14
15    def printCarInfo(getCarInfo: Car => AnyRef) {
16      for (car <- cars) println(getCarInfo(car))
17    }
18  }
19  
20  object Customer extends App {
21
22   val getOwnerInfo: (Vehicle => String) = _.owner
23   
24   Printer.printCarInfo(getOwnerInfo)
25  }

此代码取自 https://medium.com/@sinisalouc/variance-in-java-and-scala-63af925d21dc

规则是:

函数在其输入类型中是协变的,并且 其返回类型中的反变式来自Liskov替换 原理(LSP)。它说如果t支持 与U相同的操作及其所有操作要求更少(或相同) 并提供比U中相应操作更多(或相同)的 (子类型是反身的,因此S<:S)。

所以我的问题是,如果我可以传递一个需要超级类型的子类型(代码行编号15和22以上),那么为什么下面的代码不起作用呢?

class MyClass extends AnyRef
class MySubClass extends MyClass

abstract class Class {
  val f1: (Any) => Any = ???
  val f2: (Any) => Boolean = ???
  val f3: (MyClass) => Any = ???
  val f4: (MySubClass) => Boolean = ???
  val f5: (Any) => Nothing = ???
  val f6: (MyClass) => Null = ???

  val f: (MyClass) => Boolean = f4; //Error
}

更新 所以实际上,它就像给函数传递一个参数,所以我得到的参数可以是反变的[-t]返回可以是协变的[+s]

    class MyClass extends AnyRef
    class MySubClass extends MyClass

    abstract class Class {
      val f1: (Any) => Any = ???
      val f2: (MyClass) => Boolean = ???
      val f3: (MyClass) => Any = ???
      val f4: (MySubClass) => Boolean = ???
      val f5: (Any) => Nothing = ???
      val f6: (MyClass) => Null = ???

      val f: (MySubClass) => AnyVal = f2
}

这是一个有效的代码 MyClass 就像在等级制度中上升 Boolean 就像在等级制度中记录下来。