案例类是否允许使用构造函数？

发布于 2024-09-24 12:27:41 字数 839 浏览 9 评论 0原文

我有一个案例类（让我们将其命名为 Stuff），我希望能够通过扩展特征（将其称为 Marker）在运行时创建匿名子类。下面是 REPL 会话的片段，它说明了我正在尝试执行的操作：

scala> trait Marker
defined trait Marker

scala> case class Stuff(i: Int)
defined class Stuff

scala> val a = Stuff(1)
a: Stuff = Stuff(1)

scala> val b = new Stuff(1) with Marker
b: Stuff with Marker = Stuff(1)

请注意如何使用 Stuff.apply() 实例化 a，而在 b< /code> 的 case 我正在调用 case 类的构造函数。

我的问题是：使用构造函数实例化案例类是否合理？在我看来确实如此，因为案例类通常提供便利，例如 ==、< code>.equals() 和 .hashCode() 都可以。我是否遗漏了任何可能使我正在做的事情成为坏事(TM)的事情？

scala> a == b
res0: Boolean = true

scala> a.equals(b)
res1: Boolean = true

scala> a.hashCode == b.hashCode
res2: Boolean = true

原文

I have a case class (let's name it Stuff) that I want to be able to create anonymous subclasses of at run time by extending a trait (call it Marker). Here's a snippet of a REPL session that illustrates what I'm trying to do:

scala> trait Marker
defined trait Marker

scala> case class Stuff(i: Int)
defined class Stuff

scala> val a = Stuff(1)
a: Stuff = Stuff(1)

scala> val b = new Stuff(1) with Marker
b: Stuff with Marker = Stuff(1)

Note how a is instantiated using Stuff.apply(), while in b's case I'm calling the case class' constructor.

My question is: is instantiating case classes using a constructor kosher? It appears to me that it is, since the usual conveniences provided by case classes, such as ==, .equals(), and .hashCode(), all work. Am I missing anything that would brand what I'm doing a Bad Thing (TM)?

scala> a == b
res0: Boolean = true

scala> a.equals(b)
res1: Boolean = true

scala> a.hashCode == b.hashCode
res2: Boolean = true

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彼岸花似海 2024-10-01 12:27:41

以下是 Stuff.apply 的实现方式：

object Stuff {
  def apply(i: Int): Stuff = new Stuff(i)
}

因此，使用 new Stuff 没有任何坏处。

Here's how Stuff.apply is implemented:

object Stuff {
  def apply(i: Int): Stuff = new Stuff(i)
}

So there's no harm at all in using new Stuff.

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尐籹人 2024-10-01 12:27:41

对于问题

正在使用构造函数彻底实例化案例类

答案肯定是肯定的。类似的事情

val b = new Stuff(1)

根本不会造成任何问题。现在，new Stuff(1) with Marker 有所不同，因为创建了 Stuff 的匿名子类。不过，我相信这仍然没有问题。当案例类从其他案例类继承时，我所知道的问题就会出现，而您却没有这样做。但我可能不知道一些事情。

编辑：刚刚在 REPL 中测试了匹配：

scala> val b = new Stuff(1)
b: Stuff = Stuff(1)

scala> b match {case Stuff(x) => x}
res0: Int = 1

scala> b match {case Stuff(_) => true}
res1: Boolean = true

To the question

is instantiating case classes using a constructor kosher

the answer is definitely yes. Something like

val b = new Stuff(1)

poses no problems at all. Now, new Stuff(1) with Marker is different because an anonymous subclass of Stuff is created. I believe, though, that it's still unproblematic. The problems I know about appear when case classes inherit from other case classes, and you aren't doing that. But I may be unaware of something.

EDIT: just tested matching in the REPL:

scala> val b = new Stuff(1)
b: Stuff = Stuff(1)

scala> b match {case Stuff(x) => x}
res0: Int = 1

scala> b match {case Stuff(_) => true}
res1: Boolean = true

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