I have a simple question that I just can't figure out a good answer for. Why does the following Java program display 20? I would prefer a detailed response if possible please.
class Something{
public int x;
public Something(){
x=aMethod();
}
public static int aMethod(){
return 20;
}
}
class SomethingElse extends Something{
public static int aMethod(){
return 40;
}
public static void main(String[] args){
SomethingElse m;
m=new SomethingElse();
System.out.println(m.x);
}
}
Because polymorphism only applies to instance methods.
The static method aMethod invoked here
public Something(){
x=aMethod();
}
refers to the aMethod declared in Something.
The inheritance for static methods works differently then non-static one. In particular the superclass static method are NOT overridden by the subclass. The result of the static method call depends on the object class it is invoke on. Variable x is created during the Something object creation, and therefore that class (Something) static method is called to determine its value.
Consider following code:
public static void main(String[] args){
SomethingElse se = new SomethingElse();
Something sg = se;
System.out.println(se.aMethod());
System.out.println(sg.aMethod());
}
It will correctly print the 40, 20 as each object class invokes its own static method. Java documentation describes this behavior in the hiding static methods part.
Because int x is declared in the class Something. When you make the SomethingElse object, you first make a Something object (which has to set x, and it uses the aMethod() from Something instead of SomethingElse (Because you are creating a Something)). This is because aMethod() is static, and polymorphism doesn't work for static methods. Then, when you print the x from m, you print 20 since you never changed the value of x.
public class Test {
public static void main(String[] args) {
B b = new B();
b.test();
}
}
class A {
String aString = "a";
void test() {
System.out.println(aString);
}
}
class B extends A {
String aString = "b";
}
why the cant input out "b"?
i just think b.test() will call method of superClass, so b will user superClass's attribute.
why the cant input out "b"? i just think b.test() will call method of superClass, so b will user superClass's attribute.
Java class and instance variables can be inherited, but they are not virtual. That is, all variable identifiers appearing in Java code are resolved at compile time -- there is no dynamic resolution for them.
Thus, the appearance of identifier aString in method A.test() is resolved at compile time to the aString attribute of class A. This is the attribute that all invocations of that method will access, regardless of the actual class of the object on which it is invoked. If you want to use the attribute of class B when test() is invoked on an instance of that class then provide getter methods in classes A and B and have A.test() obtain the string via those.
The variable in b is hiding the variable in a. You can fix it by removing the type in b
class B extends A {
public B () {
aString = "b";
}
}
class A {
String aString = "a";
void test() {
System.out.println(aString);
}
}
class B extends A {
String aString = "b";
}
Creates 2 aString variables for B. While A will not know about B extending it, B can always access "things" from its parent class A, using the super keyword. So when you are in B, you can use aString and super.aString and they will refer the 2 different variables:
class Ideone {
public static void main (String[] args) {
A a=new A();
B b=new B();
a.test();
b.test();
b.suptest();
}
static class A {
String aString="A.aString set from A";
void test() {
System.out.println("A.test(): "+aString);
}
}
static class B extends A {
String aString="B.aString set from B";
{
super.aString="A.aString set from B";
}
void test() {
System.out.println("B.test(): "+aString);
}
void suptest() {
System.out.print("B.suptest() calling super.test(): ");
super.test();
}
}
}
You can try it on IdeOne, produces output
A.test(): A.aString set from A
B.test(): B.aString set from B
B.suptest() calling super.test(): A.test(): A.aString set from B
Where the first two lines show nothing fancy, A.aString and B.aString both contain their initial value.
But the third output line shows that super.test() call in B really "ends" in A.test(), and that the initializer block in B really altered the inherited aString field.
(Side note: the static class magic relates to the example being contained in a single file, it doesn't affect the inheritance-part)
You are correct.
Since Class A and the method test() are not abstract and the fact that the method test() was not overrided by its subClass B, the return will be exactly how was specified in Class A even if you call it from a B instance.
This question already has answers here:
Are fields initialized before constructor code is run in Java?
(5 answers)
Closed 2 years ago.
I'm trying to understand why the output of the following Java code is as like this:
class A {
public A() {
System.out.print("A()");
}
public static void main(String[] args) {
}
}
class B {
public B() {
System.out.print("B()");
}
public static void main(String[] args) {}
}
class C extends A {
B b = new B();
}
public class E05SimpleInheritance {
public static void main(String args[]) {
new C();
}
}
Output:
"A()B()"
I would imagine that when the main method of the E05SimpleInheritance public class is called the following things should happen
Non-public class C is loaded and its fields are initialized(before calling the default constructor of class C)
Since its member 'b' is an object of class B, class B is loaded in memory
Since we construct an object of Class B its constructor is called which should print B()
Default constructor of C is called which automatically calls the constructor of the superclass A which print A()
So the final output should be B()A() which is obviously wrong so I do not really understand how the code flows in this case. Can you perhaps show me why is A()B() printed instead of B()A()
Your mistake is in step 1:
Non-public class C is loaded and its fields are initialized(before calling the default constructor of class C)
This is not what’s happening. In reality, non-static fields are initialised at the beginning of the constructor, not before it. And the base class constructor is implicitly (or explicitly) invoked even before that.
In other words, javac generates code for C which is equivalent to the following:
class C extends A {
B b;
C() {
super();
b = new B();
}
}
See the code snippets below:
Code 1
public class A {
static int add(int i, int j) {
return(i + j);
}
}
public class B extends A {
public static void main(String args[]) {
short s = 9;
System.out.println(add(s, 6));
}
}
Code 2
public class A {
int add(int i, int j) {
return(i + j);
}
}
public class B extends A {
public static void main(String args[]) {
A a = new A();
short s = 9;
System.out.println(a.add(s, 6));
}
}
What is the difference between these code snippets? Both output 15 as an answer.
A static method belongs to the class itself and a non-static (aka instance) method belongs to each object that is generated from that class. If your method does something that doesn't depend on the individual characteristics of its class, make it static (it will make the program's footprint smaller). Otherwise, it should be non-static.
Example:
class Foo {
int i;
public Foo(int i) {
this.i = i;
}
public static String method1() {
return "An example string that doesn't depend on i (an instance variable)";
}
public int method2() {
return this.i + 1; // Depends on i
}
}
You can call static methods like this: Foo.method1(). If you try that with method2, it will fail. But this will work: Foo bar = new Foo(1); bar.method2();
Static methods are useful if you have only one instance (situation, circumstance) where you're going to use the method, and you don't need multiple copies (objects). For example, if you're writing a method that logs onto one and only one web site, downloads the weather data, and then returns the values, you could write it as static because you can hard code all the necessary data within the method and you're not going to have multiple instances or copies. You can then access the method statically using one of the following:
MyClass.myMethod();
this.myMethod();
myMethod();
Non-static methods are used if you're going to use your method to create multiple copies. For example, if you want to download the weather data from Boston, Miami, and Los Angeles, and if you can do so from within your method without having to individually customize the code for each separate location, you then access the method non-statically:
MyClass boston = new MyClassConstructor();
boston.myMethod("bostonURL");
MyClass miami = new MyClassConstructor();
miami.myMethod("miamiURL");
MyClass losAngeles = new MyClassConstructor();
losAngeles.myMethod("losAngelesURL");
In the above example, Java creates three separate objects and memory locations from the same method that you can individually access with the "boston", "miami", or "losAngeles" reference. You can't access any of the above statically, because MyClass.myMethod(); is a generic reference to the method, not to the individual objects that the non-static reference created.
If you run into a situation where the way you access each location, or the way the data is returned, is sufficiently different that you can't write a "one size fits all" method without jumping through a lot of hoops, you can better accomplish your goal by writing three separate static methods, one for each location.
Generally
static: no need to create object we can directly call using
ClassName.methodname()
Non Static: we need to create a object like
ClassName obj=new ClassName()
obj.methodname();
A static method belongs to the class
and a non-static method belongs to an
object of a class. That is, a
non-static method can only be called
on an object of a class that it
belongs to. A static method can
however be called both on the class as
well as an object of the class. A
static method can access only static
members. A non-static method can
access both static and non-static
members because at the time when the
static method is called, the class
might not be instantiated (if it is
called on the class itself). In the
other case, a non-static method can
only be called when the class has
already been instantiated. A static
method is shared by all instances of
the class. These are some of the basic
differences. I would also like to
point out an often ignored difference
in this context. Whenever a method is
called in C++/Java/C#, an implicit
argument (the 'this' reference) is
passed along with/without the other
parameters. In case of a static method
call, the 'this' reference is not
passed as static methods belong to a
class and hence do not have the 'this'
reference.
Reference:Static Vs Non-Static methods
Well, more technically speaking, the difference between a static method and a virtual method is the way the are linked.
A traditional "static" method like in most non OO languages gets linked/wired "statically" to its implementation at compile time. That is, if you call method Y() in program A, and link your program A with library X that implements Y(), the address of X.Y() is hardcoded to A, and you can not change that.
In OO languages like JAVA, "virtual" methods are resolved "late", at run-time, and you need to provide an instance of a class. So in, program A, to call virtual method Y(), you need to provide an instance, B.Y() for example. At runtime, every time A calls B.Y() the implementation called will depend on the instance used, so B.Y() , C.Y() etc... could all potential provide different implementations of Y() at runtime.
Why will you ever need that? Because that way you can decouple your code from the dependencies. For example, say program A is doing "draw()". With a static language, thats it, but with OO you will do B.draw() and the actual drawing will depend on the type of object B, which, at runtime, can change to square a circle etc. That way your code can draw multiple things with no need to change, even if new types of B are provided AFTER the code was written. Nifty -
A static method belongs to the class and a non-static method belongs to an object of a class.
I am giving one example how it creates difference between outputs.
public class DifferenceBetweenStaticAndNonStatic {
static int count = 0;
private int count1 = 0;
public DifferenceBetweenStaticAndNonStatic(){
count1 = count1+1;
}
public int getCount1() {
return count1;
}
public void setCount1(int count1) {
this.count1 = count1;
}
public static int countStaticPosition() {
count = count+1;
return count;
/*
* one can not use non static variables in static method.so if we will
* return count1 it will give compilation error. return count1;
*/
}
}
public class StaticNonStaticCheck {
public static void main(String[] args){
for(int i=0;i<4;i++) {
DifferenceBetweenStaticAndNonStatic p =new DifferenceBetweenStaticAndNonStatic();
System.out.println("static count position is " +DifferenceBetweenStaticAndNonStatic.count);
System.out.println("static count position is " +p.getCount1());
System.out.println("static count position is " +DifferenceBetweenStaticAndNonStatic.countStaticPosition());
System.out.println("next case: ");
System.out.println(" ");
}
}
}
Now output will be:::
static count position is 0
static count position is 1
static count position is 1
next case:
static count position is 1
static count position is 1
static count position is 2
next case:
static count position is 2
static count position is 1
static count position is 3
next case:
If your method is related to the object's characteristics, you should define it as non-static method. Otherwise, you can define your method as static, and you can use it independently from object.
Static method example
class StaticDemo
{
public static void copyArg(String str1, String str2)
{
str2 = str1;
System.out.println("First String arg is: "+str1);
System.out.println("Second String arg is: "+str2);
}
public static void main(String agrs[])
{
//StaticDemo.copyArg("XYZ", "ABC");
copyArg("XYZ", "ABC");
}
}
Output:
First String arg is: XYZ
Second String arg is: XYZ
As you can see in the above example that for calling static method, I didn’t even use an object. It can be directly called in a program or by using class name.
Non-static method example
class Test
{
public void display()
{
System.out.println("I'm non-static method");
}
public static void main(String agrs[])
{
Test obj=new Test();
obj.display();
}
}
Output:
I'm non-static method
A non-static method is always be called by using the object of class as shown in the above example.
Key Points:
How to call static methods: direct or using class name:
StaticDemo.copyArg(s1, s2);
or
copyArg(s1, s2);
How to call a non-static method: using object of the class:
Test obj = new Test();
Basic difference is non static members are declared with out using the keyword 'static'
All the static members (both variables and methods) are referred with the help of class name.
Hence the static members of class are also called as class reference members or class members..
In order to access the non static members of a class we should create reference variable .
reference variable store an object..
Simply put, from the point of view of the user, a static method either uses no variables at all or all of the variables it uses are local to the method or they are static fields. Defining a method as static gives a slight performance benefit.
Another scenario for Static method.
Yes, Static method is of the class not of the object. And when you don't want anyone to initialize the object of the class or you don't want more than one object, you need to use Private constructor and so the static method.
Here, we have private constructor and using static method we are creating a object.
Ex::
public class Demo {
private static Demo obj = null;
private Demo() {
}
public static Demo createObj() {
if(obj == null) {
obj = new Demo();
}
return obj;
}
}
Demo obj1 = Demo.createObj();
Here, Only 1 instance will be alive at a time.
- First we must know that the diff bet static and non static methods
is differ from static and non static variables :
- this code explain static method - non static method and what is the diff
public class MyClass {
static {
System.out.println("this is static routine ... ");
}
public static void foo(){
System.out.println("this is static method ");
}
public void blabla(){
System.out.println("this is non static method ");
}
public static void main(String[] args) {
/* ***************************************************************************
* 1- in static method you can implement the method inside its class like : *
* you don't have to make an object of this class to implement this method *
* MyClass.foo(); // this is correct *
* MyClass.blabla(); // this is not correct because any non static *
* method you must make an object from the class to access it like this : *
* MyClass m = new MyClass(); *
* m.blabla(); *
* ***************************************************************************/
// access static method without make an object
MyClass.foo();
MyClass m = new MyClass();
// access non static method via make object
m.blabla();
/*
access static method make a warning but the code run ok
because you don't have to make an object from MyClass
you can easily call it MyClass.foo();
*/
m.foo();
}
}
/* output of the code */
/*
this is static routine ...
this is static method
this is non static method
this is static method
*/
- this code explain static method - non static Variables and what is the diff
public class Myclass2 {
// you can declare static variable here :
// or you can write int callCount = 0;
// make the same thing
//static int callCount = 0; = int callCount = 0;
static int callCount = 0;
public void method() {
/*********************************************************************
Can i declare a static variable inside static member function in Java?
- no you can't
static int callCount = 0; // error
***********************************************************************/
/* static variable */
callCount++;
System.out.println("Calls in method (1) : " + callCount);
}
public void method2() {
int callCount2 = 0 ;
/* non static variable */
callCount2++;
System.out.println("Calls in method (2) : " + callCount2);
}
public static void main(String[] args) {
Myclass2 m = new Myclass2();
/* method (1) calls */
m.method();
m.method();
m.method();
/* method (2) calls */
m.method2();
m.method2();
m.method2();
}
}
// output
// Calls in method (1) : 1
// Calls in method (1) : 2
// Calls in method (1) : 3
// Calls in method (2) : 1
// Calls in method (2) : 1
// Calls in method (2) : 1
Sometimes, you want to have variables that are common to all objects. This is accomplished with the static modifier.
i.e. class human - number of heads (1) is static, same for all humans, however human - haircolor is variable for each human.
Notice that static vars can also be used to share information across all instances
I tried this:
class protectedfinal
{
static abstract class A
{
protected final Object a;
}
static class B extends A
{
{ a = new Integer(42); }
}
public static void main (String[] args)
{
B b = new B();
}
}
But I got this error:
protectedfinal.java:12: error: cannot assign a value to final variable a
{ a = new Integer(42); }
^
1 error
How to work around this problem?
Some people suggested here to use a constructor but this works only in some cases. It works for most objects but it is not possible to reference the object itself from within the constructor.
static abstract class X
{
protected final Object x;
X (Object x) { this.x = x; }
}
static class Y extends X
{
Y () { super (new Integer(42)); }
}
static class Z extends X
{
Z () { super (this); }
}
This is the error:
protectedfinal.java:28: error: cannot reference this before supertype constructor has been called
Z () { super (this); }
^
One could argue that it does not make much sense to store this kind of reference, because this exists already. That is right but this is a general problem which occurs with any use of this in the constructor. It is not possible to pass this to any other object to store it in the final variable.
static class Z extends X
{
Z () { super (new Any (this)); }
}
So how can I write an abstract class, which forces all child classes to have a final member which gets initialized in the child?
You have to initialize A.a in its constructor. Subclasses will use super() to pass initializer to A.a.
class protectedfinal {
static abstract class A {
protected final Object a;
protected A(Object a) {
this.a = a;
}
}
static class B extends A {
B() {
super(new Integer(42));
}
}
public static void main (String[] args) {
B b = new B();
}
}
You cannot use this until superclass constructors were called, because at this stage the object is not initialized, even Object constructor hasn't run at this point, therefore calling any instance methods would lead to unpredictable results.
In your case, you have to resolve circular reference with Z class in another way:
Z () { super (new Any (this)); }
Either use a non-final field or change class hierarchy. Your workaround with instance method super(new Any(a())); would not work for the same reason: you cannot call instance methods until superclass constructors were run.
In my personal oppinion, your problems hints towards a flaw in design.
But to answer your question. If absolutly necessary, you can change final fields in java using reflection.
And if everything fails, you can still utilize sun.misc.unsafe.
But I strongly discourage you from doing so, since it potentially kills your vm.
My work around so far is to use methods instead of final members:
class protectedfinal
{
static abstract class AA
{
protected abstract Object a();
}
static class BB extends AA
{
#Override
protected Object a() { return this; }
}
public static void main (String[] args)
{
AA a = new BB();
System.out.println (a.a());
}
}
But I would like to use final members, because I think accessing a final member is faster than calling a method. Is there any chance to implement it with final members?