Could I have some feedback on this
Given "new House("x ")" sends a string I had expected that the "House(String name)" constructor would have called the Building super class constructor "Building(String name)". In which case the answer would have been "bn x h hn x". However the answer is "b h hn x" (and yes it does run with that output).
Questions
1. Other than a call "new Building("string_value")" would there be a situation when House would call the "Building(String name)" constructor? (ie other than additional code in the House constructors?
2.Why is it that the no argument Building constructor is called, rather than the overloaded Building (String name) constructor? What I am looking is a possibility there could be many Building constructors and there could be a need to call specific super constructors from subclasses. How do you ensure which constructor (given two or more choices) is called?
Code included for ease of reference.
The answer is "b h hn x"
class Building {
Building() {System.out.print("b ");}
Building(String name) {this(); System.out.print("bn "+name);}
}
public class House extends Building {
House() {System.out.print("h ");}
House(String name) { this();System.out.print("hn "+name);}
public static void main(String a[]) {
new House("x "); }
}
Regards
Scott
If no explicit superclass constructor call is provided, and no call to a constructor in the same class is provided either, the no-args superclass constructor is always called. That's how Java is designed, and it would be too complicated and inefficient for the JVM to record which was the first constructor called and try and match it up with a superclass constructor.
If you needed to call a different superclass constructor, you would just call it explicitly, like this:
super(foo,bar);
When inheriting from another class, you must call super() in your constructor. If you don't, the compiler will insert that call for you as you can plainly see.
The superclass constructors are called because otherwise the object would be left in an uninitialized state.
Your program execution order is given below:
new House("x "); // call in main this will call same class default constructor because of this(), as you knew already that first statement must be either this() or super() if any
Call to this() in above constructor executes House() constructor. Now in House() there is no this() call so compiler puts default super() which will call base class default constructor and
hence the output is b h hn x
Related
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.
I am learning Kotlin programming language perfectly. I try to write code in different patterns and try to understand. However, I did not understand the thing. Can you help me, please?
Here it is:
open class Parent {
open val foo = 1
init {
println(foo)
}
}
class Child: Parent() {
override val foo =2
}
fun main() {
Child()
}
In this code, 0 is the output. How will this be?
This is about the order of construction — and is a subtle gotcha that's easy to fall prey to. (I'm afraid this answer is a bit long, but the issues here are well worth understanding.)
There are a few basic principles colliding here:
Superclass initialisation happens before subclass initialisation. This includes code in constructors, code in init blocks, and property initialisers: all of that happens for a superclass before any in a subclass.
A Kotlin property consists of a getter method, a setter method (if it's a var), and a backing field (if needed). This is why you can override properties; it means that the accessor method(s) are overridden.
All fields initially hold 0/false/null before they get initialised to any other value. (Normally, you wouldn't get to see that, but this is one of those rare cases. This differs from languages like C where if you don't explicitly initialise a field it can hold random values depending on what that memory was previously used for.)
From the first principle, when you call the Child() constructor, it will start off by calling the Parent() constructor. That will set the superclass's foo field to 1, and then get the foo property and print it out. After that, the Child initialisation happens, which in this case is simply setting its foo field to 2.
The gotcha here is that you effectively have two foos!
Parent defines a property called foo, and that gets accessor methods and a backing field. But Child defines its own property called foo, overriding the one in Parent — that one overrides the accessor methods, and gets its own backing field as well.
Because of that override, when the Parent's init block refers to foo, it calls the getter method which Child overrides, to get the value of Child's backing field. And that field hasn't been initialised yet! So, as mentioned above, it's still holding its initial value of 0, which is the value that the Child getter returns, and hence the value that Parent constructor prints out.
So the real problem here is that you're accessing the subclass field before it's been initialised. This question shows why that's a really bad idea! As a general rule:
A constructor/initialiser should never access a method or property that could be overridden by a subclass.
And the IDE helps you out here: if you put your code into IntelliJ, you'll see that the usage of foo is marked with the warning ‘Accessing non-final property foo in constructor’. That's telling you that this sort of problem is possible.
Of course, there are more subtle cases that an IDE might not be able to warn you about, such as if a constructor calls a non-open method that calls an open one. So care is needed.
There are occasions when you might need to break that rule — but they're very rare, and you should check very carefully that nothing can go wrong (even if someone comes along later and creates a new subclass). And you should make it very clear in comments/documentation what's going on and why it's needed.
Now, let's with java understand why. In Java, it's impossible to override fields and under the hood in Kotlin is the same. When you override a property, in fact, you override a getter, not a field. For instance, you can override a property that doesn't have a field with a property that has a field. That's totally legal. However, when both a property from a superclass and an overridden property in a subclass have fields, that might lead to unexpected results. Let's see what bytecode is generated for the Kotlin class in my example. As usual, I'll look at the corresponding Java code instead for simplicity.
public class Parent {
private final int foo = 1;
public int getFoo() {return foo;}
public Parent(){
System.out.println(getFoo());
}
}
public final class Child extends Parent {
private final int foo = 2;
public int getFoo() {return foo;}
}
public class Main
{
public static void main (String[] args) {
new Child();
}
}
Note two things here. First, the foo get to is trivial, so a field and a getter correspond to the full property. Then because the property is open and can be overridden in a subclass, its usage inside the class is compiled to a getter code, not a field code. Now, the generated code for the child class. Note that the overridden property in the parent class is also compiled to a field and a getter, and now it's another field. What happens when you create an instance of the child class? At first at the parent constructor is called, the parent constructor initializes the first fulfilled with one. But inside the init section, an overridden getter is called which calls get foo from the child class. Because the field in the child class is not yet initialized, 0 is returned. That's why 0 is printed here.
Please go through following points:
Initializer Blocks i.e. init {} block are called during an instance initialization. They are called after Primary Constructor.
In above code println(foo) is placed inside init block.
Hence, the value which gets printed i.e. 0 in this case, is the value before assignment statement open val foo = 1.
If you want the output to be 1 then make following changes:
open class Parent {
open var foo : Int = 0
init {
foo = 1
println(foo)
}
}
class Child: Parent() {
override var foo =2
}
fun main() {
Child()
}
And lastly, please go through this post. This will help you in getting better understanding of this area.
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.
How can I solve the following case?
interface I
class A(i: I)
class C : I, A(this) // << --- 'this' is not defined in this context
In short, I want to pass the class instance to super class constructor.
Is it possible in Kotlin?
P.S.
All the answers are good and technically correct. But let's give a concrete example:
interface Pilot {
fun informAboutObstacle()
}
abstract class Car(private val pilot: Pilot) {
fun drive() {
while (true) {
// ....
if (haveObstacleDetected()) {
pilot.informAboutObstacle()
}
// ....
}
}
fun break() {
// stop the car
}
}
class AutopilotCar : Pilot, Car(this) { // For example, Tesla :)
override fun informAboutObstacle() {
break() // stop the car
}
}
This example don't look too contrived, and why can't I implement it with OOP-friendly language?
No, this is not possible on the JVM. this is only available after the super class has been initialized.
From
https://docs.oracle.com/javase/specs/jvms/se9/html/jvms-4.html#jvms-4.10.2.4
The instance initialization method (§2.9.1) for class myClass sees the new uninitialized object as its this argument in local variable 0. Before that method invokes another instance initialization method of myClass or its direct superclass on this, the only operation the method can perform on this is assigning fields declared within myClass.
So the bytecode instruction aload 0 to push this on the stack is forbidden before the super-class constructor is called. That's why it cannot be passed as an argument to the super-constructor.
Kotlin was born as a JVM language and aims for maximum interoperability with Java code and a minimum overhead of its language features. While Kotlin could have chosen to orchestrate object initialization in a different way, it would create problems in mixed Java-Kotlin class hierarchies and add significant overhead.
In the good tradition of OOP languages such as Java, C# or Swift, Kotlin doesn't allow you to leak the this reference before the call to superclass initialization has completed. In your special case you're just storing the reference, but in just a slightly different case the superclass code might try to use the received object, which at that point is still uninitialized.
As a specific example of why languages don't allow this, consider a case where A is a class from a library you use and this rule is not in effect. You pass this like you do and things work fine. Later you update the library to a newer version and it happens to add something as benign as i.toString() to its constructor. It has no idea it's actually calling an overridden method on itself. Your toString() implementation observes all its invariants broken, such as uninitialized vals.
This design suffers from other problems, not just the circular initialization dependency you are struggling with now. In a nutshell, the class A expects this:
But instead you create this:
The class A has a dependency on a collaborator object of type I. It doesn't expect itself as the collaborator. This may bring about all kinds of weird bugs. For example your C.toString() may delegate to super.toString() and A.toString() (A is the super of C) may call into I.toString(), resulting in a StackOverflowError.
I can't say from your question whether A is designed for extension, which would make the C : A part correct, but you should definitely disentangle A from I.
If I have abstract class A with an init method:
abstract class A(){
init {
println("Hello")
}
}
And then class B that extends A
class B(): A()
If I instantiate B like this
fun main(args: Array<String>){
B()
}
Does the init method in A still get run and Hello gets printed?
And if not, what do I need to do to have the init method of A get run?
Yes, an init block of a base class gets run when the derived class instance is initialized.
In Kotlin, similarly to Java, an instance of a class is constructed in the following way:
An object is allocated.
The constructor of the class is called. (a)
If the class has a superclass, the superclass constructor is called before the class construction logic is executed; (i.e., the point (a) is executed recursively for the superclass, then the execution continues from here)
If the class has property initializers or init blocks, they are executed in the same order as they appear in the class body;
If the constructor has a body (i.e. it is a secondary constructor) then the body is executed.
In this description, you can see that, when B is constructed, the constructor of A is called before B initialization logic is executed, and, in particular, all init blocks of A are executed.
(runnable demo of this logic)
A small remark on terminology: init block is not actually a separate method. Instead, all init blocks together with member property initializers are compiled into the code of the constructor, so they should rather be considered a part of the constructor.