在Scala中，我如何使用多个构造函数来子类化一个Java类？ [英] In Scala, how can I subclass a Java class with multiple constructors?

问题描述

假设我有一个具有多个构造函数的Java类：

Suppose I have a Java class with multiple constructors:

class Base {
    Base(int arg1) {...};
    Base(String arg2) {...};
    Base(double arg3) {...};
}

我如何在Scala中扩展它，构造函数？在Scala中，子类只能调用它的超类的构造函数之一。

How can I extend it in Scala and still provide access to all three of Base's constructors? In Scala, a subclass can only call one of it's superclass's constructors. How can I work around this rule?

假设Java类是我不能改变的旧代码。

Assume the Java class is legacy code that I can't change.

推荐答案

很容易忘记一个trait可能会扩展一个类。如果你使用trait，你可以推迟哪个构造函数调用的决定，例如：

It's easy to forget that a trait may extend a class. If you use a trait, you can postpone the decision of which constructor to call, like this:

trait Extended extends Base {
  ...
}

object Extended {
  def apply(arg1: Int) = new Base(arg1) with Extended
  def apply(arg2: String) = new Base(arg2) with Extended
  def apply(arg3: Double) = new Base(arg3) with Extended
}

Traits自身可能没有构造函数参数，但你可以使用抽象成员来解决这个问题。

Traits may not themselves have constructor parameters, but you can work around that by using abstract members instead.

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