Derived class initialisation order - kotlin

Question

I'm currently working my way through the Kotlin docs for this section where they cover Derived class initialization order.
For the following snippet...
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")
}
when executed it prints this:
Constructing Derived("hello", "world")
Argument for Base: Hello
Initializing Base
Initializing size in Base: 5
Initializing Derived
Initializing size in Derived: 10
My question is, why when override val size: Int = (super.size + lastName.length).also { println("Initializing size in Derived: $it") }
is executed does it not print Initializing size in Base: 5 again?
I would have thought it would print something like this:
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

You're initializing Base only once.
For that reason you're also initialising size only once.
For that reason you'll execute your also block only once.
Or, to answer your question in a different manner, it doesn't print Initializing size in Base second time, because it doesn't executes name.length.also { println("Initializing size in Base: $it") } second time.