派生类初始化顺序

Question

目前，我正在研究Kotlin文档中涉及派生类初始化顺序的部分。

下面的片段..。

open class Base(val name: String) {

  init { println("Initializing Base") }

  open val size: Int = name.length.also { println("Initializing size in Base: $it") }
}

class Derived(
    name: String,
    val lastName: String
  ) : Base(name.capitalize().also { println("Argument for Base: $it") }) {

  init { println("Initializing Derived") }

  override val size: Int =
    (super.size + lastName.length).also { println("Initializing size in Derived: $it") }
}

fun main(args: Array<String>) {
  println("Constructing Derived(\"hello\", \"world\")")
  val d = Derived("hello", "world")
}

执行时，它会打印以下内容：

Constructing Derived("hello", "world") Argument for Base: Hello Initializing Base Initializing size in Base: 5 Initializing Derived Initializing size in Derived: 10

我的问题是，为什么当override val size: Int = (super.size + lastName.length).also { println("Initializing size in Derived: $it") }被执行时，它不再打印Initializing size in Base: 5？

我原以为它会印出这样的东西：

Constructing Derived("hello", "world") Argument for Base: Hello Initializing Base Initializing size in Base: 5 Initializing Derived Initializing size in Base: 5 // Print because .also is called again ? Initializing size in Derived: 10

Answer 1

您只初始化Base一次。

由于这个原因，您也只初始化size一次。

因此，您将只执行also块一次。

或者，为了以不同的方式回答您的问题，它不会第二次打印Initializing size in Base，因为它不会第二次执行name.length.also { println("Initializing size in Base: $it") }。