在Scala中不重复构造函数参数的子类化的快捷方式？ [英] Shortcut for subclassing in Scala without repeating constructor arguments?

问题描述

假设我有这样的类：

abstract class View(val writer: XMLStreamWriter) {
    // Implementation
}

class TestView(writer: XMLStreamWriter) extends View(writer) {
    // Implementation
}

View的大多数子类都不会使用不同的构造函数参数。我想要能够这样写：

Most subclasses of View are not going to take different constructor arguments. I would like to be able to write something like this:

class TestView extends View {
    // Implementation
}

有一些快捷方式来编写子类，以便您不必显式定义构造函数args并将它们传递给超类（如果我改变超类的签名，我不必重写所有的子类）

Is there some shortcut to write subclasses so that you don't have to explicitly define the constructor args and pass them to the superclass (so that I don't have to re-write all my subclasses if I change the signature of the superclass)?

推荐答案

恐怕你自己在那里。构造函数不是继承的，也不是多态和子类构造函数，而它们必须并且总是调用它们的直接超类的构造函数，不能并且不能自动完成，除非有一个零参数构造函数，超类的名称在extends子句中。

I'm afraid you're on your own there. Constructors aren't inherited or polymorphic and subclass constructors, while they must and always do invoke a constructor for their immediate superclass, do not and cannot have that done automatically, except if there's a zero-arg constructor, which is implied by the mention of the superclass's name in an "extends" clause.

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