我可以在Scala中定义“方法专用”字段吗？ [英] Can I define “method-private” fields in Scala?

问题描述

鉴于这种情况：

object ResourceManager {

  private var inited = false

  def init(config: Config) {
    if (inited)
      throw new IllegalStateException
    // do initialization
    inited = true
  }

}

有什么办法可以使 inited 以某种方式专用于init（），这样我可以确定此类中没有其他方法可以设置 inited = false ？

Is there any way that I could make inited somehow "private to init()", such that I can be sure that no other method in this class will ever be able to set inited = false?

推荐答案

来自在Scala中，如何在函数内部声明静态数据？。不要使用方法而是函数对象：

Taken from In Scala, how would you declare static data inside a function?. Don’t use a method but a function object:

val init = { // or lazy val
  var inited = false

  (config: Config) => {
      if (inited)
          throw new IllegalStateException

      inited = true
  }
}

在初始化外部范围（在 val 的情况下）或首次访问（ lazy val ），则变量的主体将被执行。因此，已初始化被设置为 false 。最后一个表达式是一个匿名函数，然后将其分配给 init 。然后，每次对 init 的进一步访问都会执行此匿名函数。

During initialisation of the outer scope (in case of val) or first access (lazy val), the body of the variable is executed. Thus, inited is set to false. The last expression is an anonymous function which is then assigned to init. Every further access to init will then execute this anonymous function.

请注意，它的行为与不完全相同之类的方法。即不带参数调用它是完全有效的。然后，它的行为就像带下划线 method _ 的方法一样，这意味着它将只返回匿名函数而不会抱怨。

Note that it does not behave exactly like a method. I.e. it is perfectly valid to call it without arguments. It will then behave like a method with trailing underscore method _, which means that it will just return the anonymous function without complaining.

如果出于某种原因您确实需要方法行为，则可以将其设为 priv val _init = ... 并从public <$ c调用它$ c> def init（config：Config）= _init（config）。

If for some reason or another, you actually need method behaviour, you could make it a private val _init = ... and call it from public def init(config: Config) = _init(config).

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