scala - 伴随对象的继承

Question

我正在尝试重构一些目前看起来像这样的模型:

case class Person(name: String, age: Int)

object Person {
    implicit val reads: Reads[Person] = (
        (JsPath \ "name").read[String] and
        (JsPath \ "age").read[Int] 
    )(Person.apply _)
}

对于看起来像这样的东西:

abstract class BaseModel {
    val pk: String  // Some stuff here that will be common
}

object BaseModel {
    implicit val reads: Reads[BaseModel] // Not sure what to do here
}

这样我就可以做到这一点:

trait MyTrait[Model <: BaseModel] {

    // More code here
    val body: JsObject = ...
    val parsed = body.validate[Model]  // Error: There is no implicit value defined for Model

}


case class NewPerson extends BaseModel {...}
object NewPerson {...} // Maybe need to extend something here


class MyController extends MyTrait[NewPerson]

<小时/>

我希望每个模型都定义一个隐式读取值，但我不确定如何在抽象类的伴随对象中指示这一点。

Answer 1

没有任何语言功能会强制抽象类/伴生对一起扩展。我通过将抽象类的“同伴”作为一个特征来克服这个缺失的环节。像这样的事情:

abstract class BaseModel {
     val pk: String
}

trait ModelCompanion[A <: BaseModel] { self: Singleton =>
    implicit val reads: Reads[A]
}

case class Person(name: String, age: Int) extends BaseModel

object Person extends BaseModel[Person] {
    ...
}

不幸的是，这仍然无法说明 MyTrait (如OP中所定义)一个Model <: BaseModel有一个同伴，其中隐式 Reads[Model]可以被找寻到。同样，需要通过要求 MyTrait 手动创建链接。保存对模型伴生对象的引用。

trait MyTrait[Model <: BaseModel] {

    def companion: ModelCompanion[Model]

    implicit def reads: Reads[Model] = companion.reads

    // More code here
    val body: JsObject = ...
    val parsed = body.validate[Model]  // Now this would work

}

object MyTraitImpl extends MyTrait[Person] {
    def companion = Person
}