I am having the following problem:
public class IntitializeTest {
static {
System.out.println("Static Initializer");
i = 1;
k = 2;
System.out.println(i+" "+k) // illegal forward reference;
}
static int i;
static int k;
{
System.out.println("Instance Initializer");
thisString = " empty ";
b = 0;
System.out.println(thisString+b) // illegal forward reference;
}
String thisString;
int b;
}
what would be the order of initialization here? for Eg: I am able to initialize before declaring the variable "b", but not able to print it because of the error.
Java docs mentions the left-hand side rule, How does this apply here.
Thanks in advance
Related
class decimaltobinary{
int y= 56;
int array[] = new int[10];
public static void main(String[] args) {
converttobinary con = new converttobinary();
}
class converttobinary{ //Error occurring at this line
for(int i = 0; i<11;i++) {
while(y > 0) {
int x = y%2;
array[i]= x;
}
}
}
}
// Error occurring at the class convert to binary.
// It says insert Class body to complete class declaration.
Well the error speaks for itself.
Insert Class body to complete class declaration.
For loops must be either inside a method or a block.
class converttobinary {
{
for (int i = 0; i < 11; i++) {
while (y > 0) {
int x = y % 2;
array[i] = x;
}
}
}
}
And then you'll have another problem in your code.
Non-static variables cannot be referenced from a static context.
When you are coding in the main method note that it is static (public static void main(...)) so either you have to make all variables static or create a new instance of decimaltobinary class.
Easiest way to solve this is by creating a new instance of decimaltobinary
converttobinary con = new decimaltobinary().new converttobinary();
Below code must be kept inside some method :
for(int i = 0; i<11;i++) {
while(y > 0) {
int x = y%2;
array[i]= x;
}
}
I am writing code in Java which has multiple methods and these methods have multiple variables. I want the other methods to access the variables of another method using actual and formal parameters. How can I do it?
I am pasting an example of the problem I'm facing.
Error : variable is not defined.
Code
public class example {
public void addition() {
int a = 0;
int b = 10;
int c = a + b;
}
public void result() {
System.out.println("The result for the above addition is" + c);
}
}
IM GETTING AN ERROR SAYING VARIABLE IS NOT DEFINED
You should declare c as global variable
public class Example {
int c;
public void addition() {
int a = 0;
int b = 10;
c = a + b;
}
public void result() {
System.out.println("The result for the above addition is " + c);
}
public static void main(String[] args) {
Example e = new Example();
e.addition();
e.result();
}
}
well, your java syntax is quite wrong... if you need to do an addition, you can do as follows:
public class Addition {
public static int addition(int a, int b)
{
int c= a + b;
return c;
}
public static void main(String[] args) {
int a = 1;
int b = 10;
int c = addition(a,b);
System.out.println("The result for the above addition is " + c);
}
}
where addition function does add a + b and return the result to your main method.
I was trying to make a program that searches for a random number, but i had problems importing the "a" variable in the other method. I would be happy if i could get some explanation. I have already tried to make the variable static, but that doesn't work
import java.util.Random;
public class verschlüsselung {
private static void nummber(int a) {
Random r = new Random();
a = r.nextInt(999);
System.out.println(a);
}
private static void search(int b) {
b = 0;
if(b =! a) {
for(b = 1; b =! a ; b++) {
if(b == a) {
System.out.println("found the number " + b);
}
}
}
}
public static void main(String args[]){
nummber(0);
search(0);
}
}
There is no such thing as using local variables in other methods.
You can return the variable from one method. Than call this method from other and get there the variable.
Declare the variable 'a' to be static and remove the parameter 'a' passed in the nummber()
function. This function does not need any input as it assigns the value of a random number to the global static variable 'a' which is accessed in the method search().
your declaration and method signature should read :
private static int a;
private static void nummber(){....}
May this help you:
private static int nummber( int a){
Random r = new Random();
a =r.nextInt(999);
System.out.println(a);
return a;
}
private static void search(int b, int a){
b = 0;
if(b =! a){
for(b =1; b =! a ; b++){
if(b == a){
System.out.println("found the number " + b);
}
}
}
}
public static void main(String args[]){
int a = nummber(0);
search(0, a);
}
If you are not allowed to initialize a final non-static data member twice, then how can I set x to something that I want in the following example?
class Temp6
{
final int x;
Temp6()
{
System.out.println(this.x);
this.x=10;
}
public static void main(String[]s)
{
Temp6 t1 = new Temp6();
System.out.println(t1.x);
}
}
Java by default gives x a value of 0, so how can I change it to 10?.
A variable marked final in Java can only be initialized once.
Simply declaring x with final int x; does not initialize it. Therefore, it is legal to assign to x in the Temp6 constructor. However, you would not be able to assign a different value to x after the constructor.
That is, the assignment to t1.x in the following:
public static void main(String[] s) {
Temp6 t1 = new Temp6();
t1.x = 11; // ERROR
}
is not legal.
Initialize final variables in the class constructor.
public class Blam
{
private final int qbert;
public Blam(int qbertValue)
{
qbert = qbertValue;
}
}
Reading this.x in your code should give an error, because final variables are not initialized upon declaration. t1.x should be 10 because x is definitely assigned at the end of the sole constructor.
You have to swap the two lines in the constructor for it to compile and it will be 10 there.
class Temp {
int x; // declaration and definition; defaulted to 0
final int y; // declaration, not initialized
Temp() {
System.out.println(x); // prints 0
x = 1;
System.out.println(x); // prints 1
x = 2; // last value, instance.x will give 2
System.out.println(y); // should be a compiler error: The blank final field y may not have been initialized
y = 3; // definite assignment, last and only value, instance.y will be 3 whereever used
System.out.println(y); // prints 3
y = 4; // compile error: The final field y may already have been assigned
}
}
I never thought about this before, interesting point here.
Final field variables behave like local variables in methods, they must be explicitly assigned before usage (definite assignment is hard to formalize, see JLS reference, but it's quite logical).
If you want to give a value to x from outside, you could do it like this:
public class Temp {
private final int x;
public Temp(int x) {
this.x = x;
}
public int getX() { return this.x; }
public static void main(String[] args) {
Temp temp = new Temp(10);
System.out.println(temp.getX()); // 10
}
}
final variable are java constants. They should be initialized before class loads.
final int x=10;
If your final variable is static then it's not like you have to give the value at the declaration itself, you can have something like -
class Demo {
static final int x;
static {
x = 10;
}
}
static block gets executed only once, at the time of class loading
How do I access the method variable having the same name as the inner class member instance or method local variable of the inner class?
class A{
int a = 10; //1
public void someMethodA(){
final int a = 20; //2
class B{
int a = 30; //3
public void someMethodB(){
int a = 40; //4
System.out.println("a = "+a); //access 4
System.out.println("a = "+this.a); //access 3
System.out.println("a = "+A.this.a); //access 1
System.out.println(?????); //how do I access value of a i.e 2
}
}
}
}
No. You can't do that. The reason being, the variable a in position 2 is a local variable, which can only be accessed with it's simple name, in the enclosing scope. From JLS §6.4:
A local variable (§14.4), formal parameter (§8.4.1), exception parameter (§14.20), and local class (§14.3) can only be referred to using a simple name (§6.2), not a qualified name (§6.6).
Now, this clears that you can only access that variable with just a. But, you have another variable in method local class B which shadows that local variable a in position 2, which is again shadowed by local variable in position 4.
Now in print statement, a would access the variable from nearmost enclosing scope, that is local variable at position 4. If you remove that variable, it will print the variable at position 3. And then if you remove it, it will print the variable at position 2.
So, the point is, there is no way to access the local variable a at position 2, because that is shadowed.
that code will never output those statements, here is an example from Sun that should explain things :
public class ShadowTest {
public int x = 0;
class FirstLevel {
public int x = 1;
void methodInFirstLevel(int x) {
System.out.println("x = " + x);
System.out.println("this.x = " + this.x);
System.out.println("ShadowTest.this.x = " + ShadowTest.this.x);
}
}
public static void main(String... args) {
ShadowTest st = new ShadowTest();
ShadowTest.FirstLevel fl = st.new FirstLevel();
fl.methodInFirstLevel(23);
}
}
I am not sure, what you actually mean, but you access a class member with the same name as a local variable with the same name like this:
public class A
{
int a = 10;
public void someMethodA()
{
int a = 5;
this.a = 20; //change the member a from 10 to 20
a = 30; // changes the local variable, which is only known in this method to 30
}
}
Typically this pattern is used in constructors, to name the params the same as member variables, for example:
class Foo
{
private int bar = 10;
private string fooBar = 20;
public Foo(int bar, string fooBar)
{
this.bar = bar;
this.fooBar = fooBar;
}
}