scala - 为什么该类型导致 "Product with Serializable with Animal"而不仅仅是 Animal？

Question

假设我有

val flag = true

为什么类型会导致 Product with Serializable with Animal而不仅仅是 Animal ?

class Animal(name : String)
case class Herbivore(name : String) extends Animal(name)
case class Carnivore(name : String) extends Animal(name)

val cow = new Herbivore("cow")
val tiger = new Carnivore("tiger")

if (flag) cow else tiger // Why is type Product with Serializable with Animal? 

Answer 1

案例类 automatically extend Product with Serializable所以

class Animal(name : String)
case class Herbivore(name : String) extends Animal(name)
case class Carnivore(name : String) extends Animal(name)

实际上是

class Animal(name : String)
case class Herbivore(name : String) extends Animal(name) with Product with Serializable
case class Carnivore(name : String) extends Animal(name) with Product with Serializable

因此最精确的类型 if (flag) cow else tiger表达式编译器可以推断为 Product with Serializable with Animal .如果我们从案例类更改为这样的类

class Animal(name : String)
class Herbivore(name : String) extends Animal(name)
class Carnivore(name : String) extends Animal(name)

那么 if-else 表达式的推断类型确实是 Animal .

根据@TravisBrown 的建议，使 ADT 根扩展 Product with Serializable像这样

abstract class Animal(name : String) extends Product with Serializable
case class Herbivore(name : String) extends Animal(name)
case class Carnivore(name : String) extends Animal(name)

将使 if-else 表达式的推断类型也 Animal .