The Kotlin documentation says that
All classes in Kotlin have a common superclass Any, that is a default super for a class with no supertypes declared
If I try and explicitly inherit from Any:
class MyClass : Any {
}
The compiler gives an error:
Kotlin: This type has a constructor, and thus must be initialized here
I haven't been able to find the documentation for the Any class. Is it possible to explicitly inherit from Any, and if so, what do you pass it?
You have to call the constructor explicitly:
class MyClass : Any()
THe constructor of Any has no parameters, thus to call it you simply provide the empty parentheses
In case we are extending a class,we need to add brackets(for implicit constructor)
class MyClass : Any()
This is similar to calling
class MyClass extends Any
{
MyClass()
{
super();
}
}
But if we are implementing an interface(interface do not have constructors),syntax should be the following
class MyClass : BaseInterface
In case when you have secondary constructor (key word constructor) you can use the next syntax
class MyClass : Any {
constructor() : super()
}
If the class has no primary constructor, then each secondary constructor has to initialize the base type using the super keyword, or to delegate to another constructor which does that.
Read more here - https://kotlinlang.org/docs/reference/classes.html
P.S. your problem could be solved easily using Android Studio feature - Project quick fix (show intention actions and quick fixes) Win - Alt + Enter, Mac - Option + Enter
Related
I want to use Kotlin delegation in a particular context.
The delegate should not be passed in the constructor.
I want to keep a reference to the delegate for later use in the code. From within the method that I override, say printMessage(), I still need to call the delegate the same way you'd call super.printMessage() in polymorphic inheritance.
I can do the first by simply instantiating an anonymous delegate in the by clause (class Derived() : Base by BaseImpl(42) using Kotlin's documentation example). However,
this prevents me from accessing the anonymous delegate, as there is no way that I know to reference it.
I want to do something similar to the following. The following however doesn't compile with error 'this' is not defined in this context.
class Derived() : Base by this.b {
val b: Base = BaseImpl(42)
override fun printMessage() {
b.printMessage()
print("abc")
}
}
I do need a separate delegate for each instance of my Derived class. So moving b as a global variable is not an option for me.
The closest I got to what I need is with an optional parameter to the constructor. This is not a good option neither, as I don't want to allow the construction of my Derived class with arbitrary delegates.
You can do this using a private primary constructor and a public secondary constructor:
class Derived private constructor(val b: Base) : Base by b {
constructor(): this(BaseImpl(42))
override fun printMessage() {
b.printMessage()
print("abc")
}
}
If you don't need a reference to the delegate, you can also say simply,
class Derived : Base by BaseImpl(42)
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.
I was studying Inheritence from Kotlin docs, and I got stuck here. There was another post which asked a question about this, but the answers were just what the docs said in a different way.
To be clear, I understood the data flow between constructors and inheritence. What I couldn't understand was how we can use an overridden property in a base class initialization. It says
It could happen directly or indirectly
What does this actually mean? How can the base class can somehow access to the overridden property in the derived class?
Also, it said
You should therefore avoid using open members in the constructors,
property initializers and init blocks.
So how can we properly use open properties?
EDIT FOR THE COMMENT:
fun main ()
{
val d = Derived("Test2")
}
open class Base()
{
open val something:String = "Test1"
init
{
println(something) //prints null
}
}
class Derived(override val something: String): Base()
What does this actually mean? How can the base class can somehow access to the overridden property in the derived class?
This is one direct way:
abstract class Base {
abstract val something: String
init {
println(something)
}
}
class Child(override val something: String): Base()
fun main() {
Child("Test") // prints null! because the property is not initialized yet
}
This prints null, which is pretty bad for a non-nullable String property.
You should therefore avoid using open members in the constructors, property initializers and init blocks.
So how can we properly use open properties?
You can use these properties in regular methods on the base class (or in custom property getters):
abstract class Base {
abstract val something: String
fun printSomething() {
println(something)
}
}
class Child(override val something: String): Base()
fun main() {
Child("Test").printSomething() // correctly prints "Test"
}
EDIT: Here are some clarifications regarding the follow-up questions in the comments.
I couldn't quite get why the code in the init block went for the parameter in the child class constructor
I think you might be confused by Kotlin's compact syntax for the primary constructors in general, which probably makes the debugger's flow hard to understand. In the Child declaration, we actually declare many things:
the argument something passed to the Child's primary constructor
the property something on the Child class, which overrides the parent's one
the call to the parent constructor (Base())
When Child() is called, it immediately calls the Base() no-arg constructor, which runs the init block.
We didn't even delegate the base constructor with a parameter or anything, but it still went for the parameter who did the overriding
You might be mixing declarations and runtime here. Although we declare things in the Base class and in the Child class, there is only 1 instance at runtime (an instance of Child) in this example code.
So, in fact, there is only 1 property called something here (only one place in memory). If the init block accesses this property, it can only be the property of the child instance. We don't need to pass anything to the Base constructor because the init block is effectively executed with the data/fields of the Child instance.
Maybe you would be less confused if you saw the Java equivalent of this. It's obvious if you think of the abstract something as a declaration of a getter getSomething(). The child class overrides this getSomething() method and declares a private something field, the getter returns the current value of the field something. But that field is only initialized after the constructor of the parent (and the init block) finished executing.
Is there any way to create an instance of Derived but not call the constructor of Base?
open class Base(p: Int)
class Derived(p: Int) : Base(p)
You actually can do it
import sun.misc.Unsafe
open class Base(p: Int){
init {
println("Base")
}
}
class Derived(p: Int) : Base(p){
init {
println("Derived")
}
}
fun main() {
val unsafe = Unsafe::class.java.getDeclaredField("theUnsafe").apply {
isAccessible = true
}.get(null) as Unsafe
val x = unsafe.allocateInstance(Derived::class.java)
println("X = $x")
}
But don't, this solution is a low-level mechanism that was designed to be used only by the core Java library and not by standard users. You will break the logic of OOP if you use it.
this is not possible. The constructor of the derived class has to call (any) constructor of the base class in order to initialise the content(fields) of the base class.
This is also the same case in Java. Just that the default constructor is called by default (if no parameters are provided in the constructor), but if you have to choose between constructors with parameters, you always have to call them explicitly, because you have to choose which values to pass into the constructor.
You must always call a constructor of a super-class to ensure that the foundation of the class is initialized. But you can work around your issue by providing a no-arg constructor in the base class. Something like this:
open class Base(p: Int?){
val p: Int? = p
constructor(): this(null)
}
class Derived(p: Int) : Base()
The way you handle which constructor of the base class is default and which parameters are nullable, etc. will depend highly on the specific case.
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?
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#.
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.
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.
open class Super {
open var name : String = "Name1"
init {
println("INIT block fired with : $name")
name = name.toUpperCase()
println(name)
}
}
class SubClass(newName : String) : Super() {
override var name : String = "Mr. $newName"
}
fun main(args: Array<String>) {
var obj = SubClass("John")
println(obj.name)
}
The above Kotlin code results in the following TypeCastException :
INIT block fired with : null
Exception in thread "main" kotlin.TypeCastException: null cannot be cast to non-null type java.lang.String
at Super.<init>(index.kt:7)
at SubClass.<init>(index.kt:13)
at IndexKt.main(index.kt:21)
As my understanding goes while inheriting from a class in Kotlin, first the primary constructors and init blocks and secondary constructors of super classes are called with passed parameters. After which the subclass can override such properties with its own version.
Then why does the above code results in the exception as described ... What am I doing wrong ... Why does the init block in super class is fired with null ...??? At first my speculation was that the init block might get fired before the actual property initialization as it is executed as a part of primary constructor but initializing the name property in the primary constructor as below gives the same error and the IDE would have warned me if so.
open class Super(open var name : String = "Name1") {
init {
println("INIT block fired with : $name")
name = name.toUpperCase()
println(name)
}
}
class SubClass(newName : String) : Super() {
override var name : String = "Mr. $newName"
}
fun main(args: Array<String>) {
var obj = SubClass("John")
println(obj.name)
}
Console :
INIT block fired with : null
Exception in thread "main" kotlin.TypeCastException: null cannot be cast to non-null type java.lang.String
at Super.<init>(index.kt:5)
at Super.<init>(index.kt:1)
at SubClass.<init>(index.kt:11)
at IndexKt.main(index.kt:19)
Am I doing something wrong here or is this a language bug...??? What can I do to avoid the error and to make the init blocks fire with the actual passed value and not null ... ??? Elaborate what is happening behind the scene. At this time I have several situations with classes like this in my actual codebase where I want to inherit from another classes, I want to maintain property names as they are...
Essentially, because you tell Kotlin that your subclass is going to be defining name now, it is not defined when the init block in Super is executed. You are deferring definition of that until the SubClass is initialized.
This behavior is documented on the Kotlin website under "Derived class initialization order":
During construction of a new instance of a derived class, the base class initialization is done as the first step (preceded only by evaluation of the arguments for the base class constructor) and thus happens before the initialization logic of the derived class is run.
...
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. Designing a base class, you should therefore avoid using open members in the constructors, property initializers, and init blocks. [emphasis mine]
FWIW, this is similar to the reason that some Java code analysis tools will complain if you refer to non-final methods in a constructor. The way around this in Kotlin is to not refer to open properties in your init blocks in the superclass.
Have the same trouble, a disgusting issue with kotlin, when subclass constructor is ignored or initialized after super class calls internal method, this is not a safe thing, if not worst i found in kotlin.