Why we should avoid using open members of base class? - kotlin

Question

While i am reading document of Kotlin, i saw that we should avoid using open properties declared at base class:
It means that, by the time of the base class constructor execution, the properties declared or overridden in the derived class are not yet initialized. If any of those properties are used in the base class initialization logic (either directly or indirectly, through another overridden open member implementation), it may lead to incorrect behavior or a runtime failure. When designing a base class, you should therefore avoid using open members in the constructors, property initializers, and init blocks.
The document said that properties in derived class are not yet initialized when base class's constructor is called. But, how can we access derived class's properties which are not initialized, from base class constructor(I assumed that the incorrect behavior or a runtime failure were caused by this situation)? Is it possible?

Answer 1

I don't know kotlin, but I'm assuming that open is the same as virtual in other languages. It is unsafe to call virtual members in a base class constructor because the base constructor is called before the derived constructor. If the overridden property requires that the derived class be fully initialized it can cause errors because the derived constructor has not yet been called when you are inside the base constructor. At least that is the way it works in .NET languages like C#.

Answer 2

Open functions in Kotlin are functions which can be overridden by a subclass. Generally, it's a good practice to limit a class's inheritance because you should provide a class with it's necessary codes to make it overridable. If your intention is not to let a class to override your base class, then you should make it final. So Kotlin make this easy by making each class and method final by default. You can find a more detailed answer in the Objects and Class chapter of the book Kotlin in Action.
The so-called fragile base class problem occurs when modifications of a base class
can cause incorrect behavior of subclasses because the changed code of the base class no
longer matches the assumptions in its subclasses. If the class doesn’t provide exact rules
for how it should be subclassed (which methods are supposed to be overridden and how),
the clients are at risk of overriding the methods in a way the author of the base class
didn’t expect. Because it’s impossible to analyze all the subclasses, the base class is
"fragile" in the sense that any change in it may lead to unexpected changes of behavior in
subclasses.
To protect against this problem, Effective Java by Joshua Bloch (Addison-Wesley,
2008), one of the best-known books on good Java programming style, recommends that
you "design and document for inheritance or else prohibit it." This means all classes and
methods that aren’t specifically intended to be overridden in subclasses need to be
explicitly marked as final .
Kotlin follows the same philosophy. Whereas Java’s classes and methods are open by
default, Kotlin’s are final by default.

Answer 3

I assume you are asking about this example in Kotlin documentation:
open class Base(val name: String) {
init { println("Initializing a base class") }
open val size: Int =
name.length.also { println("Initializing size in the base class: $it") }
}
class Derived(
name: String,
val lastName: String,
) : Base(name.replaceFirstChar { it.uppercase() }.also { println("Argument for the base class: $it") }) {
init { println("Initializing a derived class") }
override val size: Int =
(super.size + lastName.length).also { println("Initializing size in the derived class: $it")
}
}
Kotlin designers followed good practices learned, from other language mistakes, so they made class, properties, and functions closed by default for overriding or inheriting. why?
let's add the open modifier to the base class property and override it:
open class Base(open val name: String) {
init { println("Initializing a base class") }
open val size: Int =
name.length.also { println("Initializing size in the base class: $it") }
}
class Derived(
override val name: String,
val lastName: String,
) : Base(name.replaceFirstChar { it.uppercase() }.also { println("Argument for the base class: $it") }) {
init { println("Initializing a derived class") }
override val size: Int =
(super.size + lastName.length).also { println("Initializing size in the derived class: $it") }
}
fun main() {
println("Constructing the derived class(\"hello\", \"world\")")
Derived("hello", "world")
}
if you run this code the output will be like below:
Constructing the derived class("hello", "world")
Argument for the base class: Hello
Initializing a base class
**Exception in thread "main" java.lang.NullPointerException
at Base.<init> (File.kt:6)
at Derived.<init> (File.kt:12)
at FileKt.main (File.kt:23)**
The error is happening because this line of code
open val size: Int =
name.length.also { println("Initializing size in the base class: $it") }
Why? when we were trying to initialize the Derived class, first the superclass is initialized first, so the initialization is done by evaluating the super constructor argument, then the properties and init blocks in their declaration order in the class.
when it comes to val size: Int = name.length.also{...} the initialization calls the name property which is overridden by the Derived class, the one that does NOT yet initialize.
so by avoiding marking the base properties by open, you protect the base class client from abusing the class.