必须在 Scala 中覆盖 val 变量
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Must override val variable in scala
提问by zjffdu
I meet a weird problem in scala. Following is my code, class Employee extends class Person
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
But this piece of code can not been compiled, I have explicit define firstName and lastName as val variable. Why is that ? Does it mean I have to override val variable in base class ? And what is the purpose ?
但是这段代码无法编译，我已经明确定义了 firstName 和 lastName 作为 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){
}
回答by Rex Kerr
The input parameters for the constructor are not vals unless you say they are. And if they are already, why override them?
构造函数的输入参数不是 val，除非你说它们是。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) {}
If they're not vals and you want to make vals, you don't need to override:
如果它们不是 vals 并且您想创建 vals，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) {}
回答by Dirk
Since the constructor arguments have no val/var declaration in Person, and as Person is no case class, the arguments will not be members of class Person, merely constructor arguments. The compiler is telling you essentially: hey, you said, that firstName and lastName are members, which override/redefine something inherited from a base class - but there is nothing as far as I can tell...
由于构造函数参数在 Person 中没有 val/var 声明，并且由于 Person 不是 case 类，因此参数将不是类 Person 的成员，而只是构造函数参数。编译器本质上是在告诉你：嘿，你说的 firstName 和 lastName 是成员，它们覆盖/重新定义从基类继承的东西 - 但据我所知，没有什么......
class Person(val firstName: String, val lastName: String)
class Employee(fn: String, ln: String, val salary: BigDecimal) extends Person(fn, ln)
You do not need to declare firstName/lastName as overrides here, btw. Simply forwarding the values to the base class' constructor will do the trick.
顺便说一句，您不需要在这里将 firstName/lastName 声明为覆盖。只需将值转发到基类的构造函数即可。
回答by Sam Stainsby
You might also consider redesigning your super classes as traits as much as possible. Example:
您也可以考虑将您的超类重新设计为尽可能多的特征。例子：
trait Person {
def firstName: String
def lastName: String
}
class Employee(
val firstName: String,
val lastName: String,
val department: String
) extends Person
or even
甚至
trait Employee extends Person {
def department: String
}
class SimpleEmployee(
val firstName: String,
val lastName: String,
val department: String
) extends Employee
回答by Synesso
Unless I've misunderstood your intention, here's how to extend Person.
除非我误解了您的意图，否则这里是如何扩展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
