错误或特征：Kotlin允许在继承中将'val'更改为'var' [英] Bug or Feature: Kotlin allows to change 'val' to 'var' in inheritance

问题描述

我刚刚开始探索Kotlin语言。我正在努力继承遗产，变量和副作用。

I just started to explore the language Kotlin. I'm struggling with inheritance, var&val and side-effects.

如果我宣布一个特质 A a val x 并在 AImpl 中覆盖 x 这是可能的将其覆盖为 var （参见下面的代码）。令人惊讶的是， A 中的 print（）方法受到 x 即使 x 是 A 中的值。这是一个错误还是一个功能？

If I declare a trait A with a val x and override x in AImpl it is possible to override it as var (see code below). Surprisingly the print() method in A is affected by the reassignment of x even though x is a value in A. Is this a bug or a feature?

代码：

trait A {
  fun print() {
    println("A.x = $x")
  }

  val x : Int;
}

class AImpl(x : Int) : A {
  override var x = x; // seems like x can be overriden as `var`
}

fun main(args: Array<String>) {

  val a = AImpl(2)

  a.print() // A.x = 2

  a.x = 3; // x can be changed

  // even though print() is defined in trait A
  // where x is val it prints x = 3
  a.print() // A.x = 3

}

我知道这个事实如果我明确定义 a 类型 A ，则不允许更改 x ：

I'm aware of the fact that if I define a with type A explicitly it is not allowed to change x:

val a = AImpl(2) : A
a.x = 3 // ERROR: value x cannot be reassigned

但是，正如第一个案例所示，继承会导致副作用，但显然不是打算在 A 中。如何保护值不被继承更改？

But as the first case shows, inheritance can cause side effects which are clearly not intended in A. How do I protect values from being changed by inheritance?

推荐答案

您可以使 val final ，即完全禁止覆盖它。
如果在类中定义 val ，默认情况下它是 final 。

You can make your val final, i.e. forbid overriding it at all. If you define a val in a class, it is final by default.

另外，如果你需要用 var val $ c>，但不希望setter公开，你可以这么说：

Also, if you need to override a val with a var, but do not want the setter to be public, you can say so:

override var x = 1
    private set

覆盖 val var 是一项功能。它相当于添加set方法，而在超类中只有get-method。这对于实现某些模式非常重要，例如只读接口。

Overriding a val with a var is a feature. It is equivalent to adding a set-method while in the superclass there was only a get-method. And this is rather important in implementing some patterns, such as read-only interfaces.

没有办法保护你的 val 从允许改变突变的方式被覆盖而不是使 final ，因为 val 不意思是不可变引用，但仅仅是只读属性。换句话说，当您的特征 A 声明 val 时，表示通过类型<的引用code> A 客户端无法写入此 val ，没有其他保证，或者确实可能。

There's no way to "protect" your val from being overridden in a way that allows changing mutation other than making it final, because val does not mean "immutable reference", but merely "read-only property". In other words, when your trait A declares a val, it means that through a reference of type A the client can not write this val, no other guarantees intended, or indeed possible.

P.S。分号在Kotlin中是可选的，可以完全省略它们

P.S. Semicolons are optional in Kotlin, feel free to omit them altogether

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