kotlin - 派生类初始化顺序

Question

目前，我正在本节的Kotlin文档中介绍Derived class initialization order。

对于以下代码段...

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")
}

执行时将输出以下内容:
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，它不会再次打印Initializing size in Base: 5吗？

我本以为它会打印如下内容:
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

您只初始化一次Base。
因此，您也只初始化一次size。
因此，您将只执行一次also块。

或者，以另一种方式回答您的问题，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") }。