Question

我在scala中遇到了一个奇怪的问题。以下是我的代码，类Employee扩展类Person

但是这段代码无法编译，我已经明确地将firstName和lastName定义为val变量。这是为什么？这是否意味着我必须覆盖基类中的val变量？目的是什么？

class Person( firstName: String,  lastName: String) {

}

class Employee(override val firstName: String, override val lastName: String, val depart: String)
    extends Person(firstName,lastName){

}

Answer 1

构造函数的输入参数不是val，除非你说它们。如果它们已经存在，为什么要覆盖它们？

class Person(val firstName: String, val lastName: String) {}
class Strange(
  override val firstName: String, override val lastName: String
) extends Person("John","Doe") {}
class Employee(fn: String, ln: String, val depart: String) extends Person(fn,ln) {}

如果它们不是val并且您想要制作val，则不需要覆盖：

class Person(firstName: String, lastName: String) {}
class Employee(
  val firstName: String, val lastName: String, val depart: String
) extends Person(firstName,lastName) {}

Answer 2

由于构造函数参数在Person中没有val / var声明，并且Person不是case类，因此参数不是Person类的成员，只是构造函数参数。编译器基本上告诉你：嘿，你说，firstName和lastName是成员，它覆盖/重新定义从基类继承的东西 - 但据我所知，没有任何东西......

class Person(val firstName: String, val lastName: String)
class Employee(fn: String, ln: String, val salary: BigDecimal) extends Person(fn, ln)

你不需要在这里声明firstName / lastName作为覆盖，顺便说一下。只需将值转发给基类的构造函数即可。

Answer 3

您也可以考虑尽可能多地将超类重新设计为特征。例如：

trait Person {
  def firstName: String
  def lastName: String
}

class Employee(
  val firstName: String,
  val lastName: String,
  val department: String
) extends Person

甚至

trait Employee extends Person {
  def department: String
}

class SimpleEmployee(
  val firstName: String,
  val lastName: String,
  val department: String
) extends Employee

Answer 4

除非我误解了你的意图，否则以下是如何延长Person。

Welcome to Scala version 2.8.0.final (Java HotSpot(TM) Client VM, Java 1.6.0_21).
Type in expressions to have them evaluated.
Type :help for more information.

scala> class Person( firstName: String, lastName: String)
defined class Person

scala> class Employee(firstName: String, lastName: String, depart: String) extends Person(firstName, lastName)
defined class Employee

必须覆盖scala中的val变量

4 个答案: