I have written a simple code with a super class and two sub classes in order to track order or initialization. My static initializers behave like expected and initialize in the proper order. But my instance initializers seem to run twice before my constructors. I know that's the order of initialization but I'm trying to gain some insight an follow the flow and am lost as to why the instance initializers print twice. Here's my code and the output I receive. (see below)
class Bird{
static {System.out.println("Static Initializer 1");}
{System.out.println("Instance Initializer 1");}
int feathers = 0;
Bird(int x){ this.feathers = x; }
Bird fly() { return new Bird(1); }
}
class Parrot extends Bird {
static {System.out.println("Static Initializer 2");}
{System.out.println("Instance Initializer 2");}
protected Parrot(int y){ super(y); }
protected Parrot fly(){ return new Parrot(2); }
}
public class Macaw extends Parrot {
static {System.out.println("Static Initializer 3");}
{System.out.println("Instance Initializer 3");}
public Macaw(int z){ super(z); }
public Macaw fly(){ return new Macaw(3); }
public static void main(String... args){
Bird p = new Macaw(4);
System.out.println(((Parrot)p.fly()).feathers);
}
}
Results:
Probably because your fly() method literally creates a new instance:
public Macaw(int z){ super(z); }
public Macaw fly(){ return new Macaw(3); } <---- NEW
public static void main(String... args){
Bird p = new Macaw(4); // Instance 1
System.out.println(((Parrot)p.fly()).feathers); // Calling fly() creates instance 2
}
Related
I am just learning Java concepts.
Can anyone let me know why i am not able to run this program?
package innerClasses;
public class Test {
int i=10;
static int j=20;
public void m1() {
int k=30;
final int m=40;
class Inner {
public void m2() {
System.out.println(i);
}
}
}
public static void main(String[] args) {
Test t = new Test();
Test.Inner in = t.new Inner();
t.m1();
}
}
Can anyone let me know why i am not able to run this program?
The most basic reason is because of scope. In order to do
Test.Inner in = t.new Inner();
Inner must be defined in Test, but it is instead defined in m1 scope.
The class Inner is declared inside the method m1(), what makes it not available outside this method.
Your code has to look like the following to be able to run, although it will not print anything...
public class Test {
int i=10;
static int j=20;
public void m1() {
int k=30;
final int m=40;
}
class Inner {
public void m2() {
System.out.println(i);
}
}
public static void main(String[] args) {
Test t = new Test();
Test.Inner in = t.new Inner();
t.m1();
}
}
Replacing t.m1(); by in.m2(); will output 10.
EDIT
In case you have to create the inner class inside the method, make it like
public class Test {
int i=10;
static int j=20;
public void m1() {
int k=30;
final int m=40;
class Inner {
public void m2() {
System.out.println(i);
}
}
// this is what makes it run
Inner myInner = new Inner();
myInner.m2();
}
public static void main(String[] args) {
Test t = new Test();
t.m1();
}
}
to compile and run.
IMHO this is not a good way to go...
A method inner class is only visible to that method only so you can't use this at any other location.
for using this class you have to declare it outside of the method.
You cannot compile it as the scope of Inner class is the m1 method.
If you want to be able to create instances of Inner class you can define it directly inside the Test:
public class Test {
int i=10;
static int j=20;
public void m1() {
int k=30;
final int m=40;
}
// On the Test class level
class Inner {
public void m2() {
System.out.println(i);
}
}
public static void main(String[] args) {
Test t = new Test();
Test.Inner in = t.new Inner();
t.m1();
in.m2(); // prints 10
}
}
Another feature provided by Java is to use anonymous classes. It can be created by implementation of some Interface:
// Define an interface to be able to implement it inside the method
interface Inner {
void m2();
}
public class Test {
int i=10;
static int j=20;
public void m1() {
int k=30;
final int m=40;
// Implement interface and call method on it
// Functional style:
((Inner) () -> System.out.println(i)).m2();
// Legacy style:
new Inner() {
#Override
public void m2() {
System.out.println(i);
}
}.m2();
}
public static void main(String[] args) {
Test t = new Test();
t.m1(); // prints 10 twice
}
}
or extending some class:
// A class we going to extend
class Inner {
void m2() {
System.out.println(11);
}
}
public class Test {
int i=10;
static int j=20;
public void m1() {
int k=30;
final int m=40;
// Extend inner class and call method on it
new Inner() {
void m2() {
System.out.println(i);
}
}.m2();
}
public static void main(String[] args) {
Test t = new Test();
t.m1(); // prints 10, not 11
}
}
So the best way for you depends on what code design do you want to get finally.
Your class Inner is what the JLS calls a Local Class (14.3. Local Class Declarations).
The scope of a Local Class is defined as (6.3. Scope of a Declaration):
The scope of a local class declaration immediately enclosed by a block (§14.2) is the rest of the immediately enclosing block, including its own class declaration.
The scope of a local class declaration immediately enclosed by a switch block statement group (§14.11) is the rest of the immediately enclosing switch block statement group, including its own class declaration.
In your case it is declared in a block, which is the body of your method. So the scope where your class is visible is the rest of this method body.
As the type Inner is not visible in main(), you cannot use it there. You could create instances of it and use them within m1(), though.
The very basic reason for compile time error is "Scope".
As per your code, class Inner is defined inside method m1 (this class is called Local class/method local class),so if you observe the scope of method variable, its within the declared method only and we cannot access any method variable outside that method and this is reason, the scope of class Inner is limited to the m1 method
so if you want to instantiate class Inner and invoke its methods then you must do it in m1 method (in code , you are trying to create class Inner object outside method m1, which is not possible) hence the code would be
public class Test {
int i = 10;
static int j = 20;
public void m1() {
int k = 30;
final int m = 40;
class Inner {
public void m2() {
System.out.println(i);
}
}
Inner in = new Inner();
in.m2();
}
public static void main(String[] args) {
Test t = new Test();
t.m1();
}
}
some more info , the local classes can be used when any repeated functionality is required inside a method (as nested methods are not allowed in java, so we can create inner class) and off course if we are not interested to create class level method for example
class Outer {
public void cal() {
class Inner {
public void sum(int x, int y) {
System.out.println("sum : "+(x+y));
}
}
Inner i= new Inner();
i.sum(10, 20); //sum is repeatdly needed in between code
i.sum(100, 200);
i.sum(1000, 2000);
i.sum(10000, 20000);
}
}
public class TestClass {
public static void main(String[] args) {
new Outer().cal();
}
}
I have a code-
public class Hello
{
void create()
{
Inner obj=new Inner();
obj.r=100; //Able to access private variable x
obj.display(); //displays 100
}
class Inner
{
private int r=45;
void display()
{
System.out.println("r is : "+r);
}
}
public static void main(String[] args)
{
Hello ob=new Hello();
ob.create();
}
}
In the above code,by creating an instance of the inner class,we are able to access the private variable defined in that class.But this is not in the case of inheritance.Why it is so?For e.g.,in this code-
class One
{
private int x;
void getData()
{
x=10;
}
void display()
{
System.out.println("x is : "+x);
}
}
class Two extends One
{
int y;
void putData()
{
One o=new One();
o.x=13; //Error
}
}
public class File
{
public static void main(String[] args)
{
Two to=new Two();
to.putData();
}
}
What is the exact reason behind it?Thanks in advance...
See the Java Language Specification.
Otherwise, if the member or constructor is declared private, then access is permitted if and only if it occurs within the body of the top level class (§7.6) that encloses the declaration of the member or constructor.
Meaning that a top-level class can access the private members of it's nested classes.
Or said another way: Private means private to the top-level class and all it's nested classes, not private to the nested class itself.
This program is supposed to individually call the funFact of each subclass but instead, it calls the funFact method only from the Mammal class. What am I doing wrong?
public class MammalFacts{
public static class Mammal{
public static String funFact(){
return "If you are reading this, there's a 70% chance you're a mammal";
}//end funFact
}//end mammal
public static class Primate extends Mammal{
public static String funFact(){
return "All primates can fly";
}
}//end Primate
public static class Monkey extends Primate{
public static String funFact(){
return "Monkies will rule the earth someday";
}
}
public static void main(String[]args){
Mammal[]i = new Mammal[3];
i[0] = new Mammal();
i[1] = new Primate();
i[2] = new Monkey();
for(int c = 0; c < i.length; c++){
System.out.println(i[c].funFact());
}
}//end of main
}//end MammalFacts
funFact is static. Overriding doesn't work on static methods.
Remove the static keyword from all your methods (as you are calling them via an instance reference anyway) and it will work as you expected.
If you "override" the static method, you are hiding the method, not really overriding.
read this:
https://docs.oracle.com/javase/tutorial/java/IandI/override.html
the "Static Methods" section exactly answers your question.
Why doesn't a static block in the class get executed when I don't create a reference variable for an object (anonymous) of that class?
For example, let us consider this simple class:
public class StaticDemo {
private static int x;
public static void display(){
System.out.println("Static Method: x = "+x);
}
static {
System.out.println("Static Block inside class");
}
public StaticDemo(){
System.out.println("Object created.");
}
}
Another class using it:
public class UseStaticDemo {
public static void main(String[] args) {
StaticDemo Obj = new StaticDemo();
Obj.display();
System.out.println("------------");
new StaticDemo().display();
}
}
Output:
Static Block inside class
Object created.
Static Method: x = 0
------------
Object created.
Static Method: x = 0
A static initializer block only runs once, when the class is loaded and initialized.
Also, static methods have absolutely no relation to any instances. Doing
new StaticDemo().display();
is pointless and unclear.
I am trying to understand my way around polymorphism in Java. I created a parent class that has too many common methods that all children will use in the same manner.
Each of the subclasses' children all share static information, These variables or information will be used in the methods declared only in the parent.
The problem wish accessing static variables from Parent methods seems not really possible,
Its a solution to declare the common information per instance but since there will be 1000s of instances its such a waste of memory.
A simple elaboration of what i mean is the following code :
class testParent {
static int k;
public void print()
{
System.out.println(k);
}
}
class testChild2 extends testParent
{
static
{
testChild2.k =2;
}
}
public class testChild1 extends testParent{
static
{
testChild1.k = 1;
}
public static void main(String[] args)
{
new testChild1().print();
new testChild2().print();
new testChild1().print();
}
}
the output i expect was
1
2
1.
but what happens is :
1
2
2
One might think that on the initiation of each subclass the static variables of this subclass is set and then all methods referring to this subclass has access to the corresponding 'k' value.
But what actually happens is that all subclasses edit in the same static variable that is shared along all subclasses and hence destroys my whole point of using static variables for each subclass and its instances and using commmon methods in the Parent accessing these variables.
Any idea how can this be done ?
An option is to access the subclasses' static data through an abstract (non-static) method:
abstract public class Parent {
protected abstract int getK();
public void print() {
System.out.println(getK());
}
}
public class Child1 extends Parent {
private static final int CHILD1_K = 1;
protected int getK() { return CHILD1_K; }
}
public class Child2 extends Parent {
private static final int CHILD2_K = 2;
protected int getK() { return CHILD2_K; }
}
When you make new testChild2().print(); the static block on testChield2 was executed and change the value to 2.
static blocks only execute once when loaded by the ClassLoader.
This one give the output you want:
class testParent {
static int k;
public void print()
{
System.out.println(k);
}
}
class testChild2 extends testParent
{
{
testChild2.k =2;
}
}
public class testChild1 extends testParent{
{
testChild1.k = 1;
}
public static void main(String[] args)
{
new testChild1().print();
new testChild2().print();
new testChild1().print();
}
}
Non static code blocks execute everytime the class is instanciated.
Premature optimization is the root of all evil. I don't think you'll run into any memory issues with thousands of instances, each with their own data, unless you're working on a tiny embedded system of some kind. Static variables are not intended to do what you're trying to do with them.
Static variables are specific to the class itself. If you want the same field in different instances of a class to have different values, then that field cannot be static.
The solution: don't make k static.
class testParent {
int k;
public void print()
{
System.out.println(k);
}
}
class testChild2 extends testParent
{
{
this.k =2;
}
}
class testChild1 extends testParent{
{
this.k = 1;
}
public static void main(String[] args){
new testChild1().print();
new testChild2().print();
new testChild1().print();
}
}
Demo
(ignore the static class business - that's just to make it work in ideone).