Here is the code that keeps saying illegal start of expression:
public static conversionRate= 4.546;
Here is the full code:
/**
* Write a description of class VolumeConversion here.
*
* #author (Aneeqa Rustam)
* #version (07/08/2014)
*/
public class VolumeConversion
{
// instance variables - replace the example below with your own
/**
* Constructor for objects of class VolumeConversion
*/
public VolumeConversion()
{
public static conversionRate= 4.546;
znaslcmlkmlskm(String[]args)
//Declare the variable and constants
double litres= 0;
double gallon= 14;
//Perform the conversion calculation
litres= gallon* conversionRate;
//This is the output result that is going to be shown to the user
System.out.println("The number of litres in "+gallons+ "gallons is" +litres);
}
}
You need a type for the variable. For example:
public static float conversionRate = 4.546f;
You also want to place that outside of the constructor, as a class level variable.
Type is missing in the variable declaration
The variable conversionRate doesn't have a type in its declaration.
Possible solutions:
public static float conversionRate = 4.546f;
public static double conversionRate = 4.546;
Besides that you try to declare this variable in the constructor (a "method"). That does not work. It has to be declared within class and not in methods.
Well, this one is quite obvious - you haven't defined the datatype for your conversionRate variable. What you'd probably want to use here is the double datatype, but I'd also suggest looking into BigDecimals for further reference.
Code sample:
public static double CONVERSION_RATE = 4.546;
I would personally recommend against using the float datatype (as #MrTux recommended) in real life projects, as it tends to make your code clumsier (unnecessary casting & parsing) and has the obvious limitation of a restricted value range. The performance penalty that results from the use of double in place of float, however, is miniscule in most cases.
Related
Non-static final variables can be assigned a value only once.
But why this assignment can happen only either within a declaration or in the constructor?
A final variable is defined to be immutable i.e., the value assigned to it will be the one and only for that variable x. As one can read from the JLS(§4.12.4.)
A variable can be declared final. A final variable may only be
assigned to once.
Now the constructor is just like any other method, except that it is the one that gets executed first when an object (non-static) is created from the class.
Hence, final variables can be assigned through constructors.
For example take the following code:
public class Test {
public final int x;
public Test(int x) {
this.x = x;
}
}
Compiler accepts this invocation because it is guaranteed that for that particular object its class's constructor gets invoked first and doesn't invoked again (i.e. constructor gets invoked one and only one time during the entire lifetime of object.)
However following code throws error: Non-static field 'x' cannot be referenced from a static context
public class Test {
public final int x;
static {
x = 5;
}
public Test(int x) {
this.x = x;
}
}
Since x is not a static field, it cannot be initiated within a static block.
This code would also throw error: Cannot assign a value to final variable 'x'
public class Test {
public final int x;
public Test(int x) {
this.x = x;
}
public void setX(int x) {
this.x = x;
}
}
That is because it is not guaranteed for this object, that the method setX would run first and only once. The programmer could call this method multiple times. Hence, the compiler throws an error.
So there is no way to make a variable "initializable" only once (e.g.,
a setter would block if variable was already assigned before) solely
with java syntax? I thought final might work this way but now I see
it's not.
For your question, you could simply make a variable private and add the condition to the setter method to add value only if variable is null.
For example:
public class Test {
private Integer x;
public Test() {
}
public Test(int x) {
this.x = x;
}
public void setX(int x) {
if (null == this.x) this.x = x;
}
public static void main(String[] args) {
Test y = new Test(5);
System.out.println(y.x);
y.setX(20);
System.out.println(y.x);
}
}
This is not thread safe by the way. I just added a simple example.
What does the keyword final mean in Java?
When used in a class declaration, it means that the class cannot be extended.
When used in a method, it means that the method cannot be overridden.
When used in a method parameter, it means the value of such parameter cannot be changed inside the method. (local constant)
When used in a class field ("variable), it means that it is a global constant.
Values for constants must be resolved at compile time. And, as the word implies, constants fields cannot change value. Therefore, the compiler does not allow the value to be set from a setter (mutator) method.
Contrary to what many believe, for a field to be constant, it does not have to be declared static and final. That said, since the value of a constant cannot be changed, each class instance will share the same value. Therefore, explicitly making them static reenforces this notion.
There is a fifth use of the keyword final and this is when used when local variables are declared. This is a more lengthy explanation.
What happens when you compile code?
I updated my answer because I think part of the problem is that some developers don't quite understand what happens when the code is compiled. As I mentioned before, constant values are resolved at COMPILE TIME. To understand this concept, consider the following example:
public class MyClass {
private final double PI = 3.14159;
// rest of class left out intentionally
}
If I compile this class on my laptop and then I deploy the code to some remote server, how does the server know that the global constant field PI has an assigned value of 3.14159? This is because when I compile this code, this value gets packaged with the byte code. The class constructor doesn't come into play at all in this case. HOWEVER, if the constant field is initialized to its DEFAULT value, then permanent (constant) value may be assigned via the constructor
public class MyClass {
private final double PI; // default value of 0.0
public MyClass(double value) {
PI = value;
}
// rest of code omitted intentionally
}
Here's where declaring a constant as static makes a difference. If a constant is also static, you can't do the above because calling a constructor implies that you can have multiple instances of MyClass and each instance could set a different value. This is clearly a violation of what a static member is. So, if you MUST declare a field as both static and final, understand that you cannot assign a value using this second approach. Only the first one I showed is allowed.
Final Stop's a Variable’s Reassignment
a short simple answer:
Use the keyword final when you want the compiler to prevent a variable from being re-assigned to a different object.
Whether the variable is a static variable, member variable, local variable, or argument/parameter variable, the effect is entirely the same.
Hope this helps friend =)
#StaySafe
For some background, I'm currently on chapter 8 in my book, we finished talking about arraylists, arrays, if statements, loops etc. Now this part of the book talks about call by reference,value and some other pretty neat things that seem odd to me at first.I've read What situation to use static and some other SO questions, and learned quite a bit as well.
Consider the following example my book gave (among many examples)
There is another reason why static methods are sometimes necessary. If
a method manipulates a class that you do not own, you cannot add it to
that class. Consider a method that computes the area of a rectangle.
The Rectangle class in the standard library has no such feature, and
we cannot modify that class. A static method solves this problem:
public class Geometry
{
public static double area(Rectangle rect)
{
return rect.getWidth() * rect.getHeight();
}
// More geometry methods can be added here.
}
Now we can tell you why the main method is static. When the program
starts, there aren’t any objects. Therefore, the first method in the
program must be a static method.
Ok, thats pretty cool, up until now I've just been really blindly putting public in front of all my methods, so this is great to know. But the review small problem on the next page caught my attention
The following method computes the average of an array list of numbers:
public static double average(ArrayList<Double> values)
Why must it be a static method?
Here I was like wait a sec. I'm pretty sure I did this without using static before. So I tried doing this again and pretty easily came up with the following
import java.util.ArrayList;
class ArrList
{
private double sum;
public ArrList()
{
sum = 0;
}
public double average(ArrayList <Double> values)
{
for(int i = 0; i < values.size() ; i++)
{
sum+=values.get(i);
}
return sum / values.size();
}
}
public class Average
{
public static void main(String [] args)
{
ArrList arrListObj = new ArrList();
ArrayList<Double> testArrList = new ArrayList<Double>();
testArrList.add(10.0);
testArrList.add(50.0);
testArrList.add(20.0);
testArrList.add(20.0);
System.out.println(arrListObj.average(testArrList));
}
}
TLDR
Why does my book say that public static double average(ArrayList<Double> values) needs to be static?
ATTEMPT AT USING STATIC
public class Average
{
public static void main(String [] args)
{
ArrayList<Double> testArrList = new ArrayList<Double>();
ArrayList<Double> testArrListTwo = new ArrayList<Double>();
testArrList.add(10.0);
testArrList.add(50.0);
testArrList.add(20.0);
testArrList.add(20.0);
testArrListTwo.add(20.0);
testArrListTwo.add(20.0);
testArrListTwo.add(20.0);
System.out.println(ArrList.average(testArrList));
System.out.println(ArrList.average(testArrListTwo)); // we don't get 20, we get 53.3333!
}
}
It doesn't.
The only method which needs to be static is the initial main() method. Anything and everything else is up to you as the programmer to decide what makes sense in your design.
static has nothing to do with public accessors (as you allude to), and it has nothing to do with the technical operation being performed. It has everything to do with the semantics of the operation and the class which holds it.
An instance (non-static) method exists on a particular instance of a class. Semantically it should perform operations related to that specific instance. A static method exists on a class in general and is more conceptual. It doesn't do anything to a particular instance (unless it's provided an instance of something as a method argument of course).
So you really just need to ask yourself about the semantics of the operation. Should you need new instance of an object to perform an operation? Or should the operation be available without an instance? That depends on the operation, on what the objects represent, etc.
If it is not static, then any other class that wants to use this method must first create an instance of this object.
From some other class:
Average.average(new ArrayList<Double>()); // legal only if static
new Average().average(new ArrayList<Double>()); // necessary if not static
// and only makes sense if Average can be instantiated in the first place
It's legal to make it an instance (i.e. not static) variable, but the method is actually harder to understand. If it is static then whoever reads the code knows it does not use any member variables of the class.
// In the class body
int x = 0; // member variable
public static double average() {
x = x + 1; // illegal
}
The less something can do, the easier to understand what it does do.
Static methods like the area, average are usually utility functions. You don't need any object to use an utility function. For example consider Math.pow you don't need to instantiate any object to use the power function, just use Math.pow(10.0, 2.0) to get (10.0)^2
In short :
Static method means class method, that is no instance of that object is needed to invoke.
whereas your average method is an instance method, you need an object to invoke that method.
I have trouble understanding why double xx and yy are put final in the constructor. Why don't I just put them double xx and double yy. Why
do they have to be final? I guess the whole purpose of this is
creating an immutable object.
public class Point {
private final double x, y;
private double distance;
public Point(final double xx, final double yy) {
this.x = xx;
this.y = yy;
this.distance = -1;
}
}
There is no need for these parameters to be final.
There are two reasons to make parameters final:
To make use of them in an inner class declared in that function;
To prevent their values from being changed accidentally.
Clearly (1) doesn't apply.
(2) isn't necessary because it's such simple code, and you can see that it's not changing the parameters.
There is a school of thought which says that all parameters and local variables should be declared final as a matter of course, as it makes it easier to reason about the code, in the same way that using immutable types makes it easier to reason about code using them.
There is another school of thought which says that adding final everywhere is just unnecessary noise, and, if you are writing methods where you can't tell if the value is changing, your methods are too long.
Largely, making parameters and local variables final comes down to personal/team preference.
Declaring x and y variables as final solves the purpose (i.e. The value must not be changed later). There is no point in declaring the constructor's parameters as final.
I've got this problem, I have a java file that obtains 2 variables from another file and is supposed to add them together and return the summed value. So far it works on obtaining the values aFirst and aSecond but I'm not sure why value one and two is lost (is back at 0) when it gets to the sum method. This is for an assignment I have for homework.
public class Pair
{
private double one, two ;
public Pair(double aFirst, double aSecond)
{
double one = aFirst;
double two = aSecond;
}
public double sum()
{
double xys = one + two;
return(xys);
}
}
You're declaring one and two as local variables, shadowing the instance variables.
the problem is the constructor. you are creating local variables and not using the class fields
private double one, two ;
public Pair(double aFirst, double aSecond)
{
this.one = aFirst;
this.two = aSecond;
}
you can do it without the "this." but dont put type ahead
Use this to access the class member variables in override cases when you have the same variable name in local scope and in the class members.
public class Main {
static final int alex=getc();
static final int alex1=Integer.parseInt("10");
static final int alex2=getc();
public static int getc(){
return alex1;
}
public static void main(String[] args) {
final Main m = new Main();
System.out.println(alex+" "+alex1 +" "+alex2);
}
}
Can someone tell me why this prints: 0 10 10? I understand that it's a static final variable and its value shouldn't change but it`s a little difficult to understand how the compiler initializes the fields.
It's an ordering problem. Static fields are initialized in the order that they are encountered, so when you call getc() to inititalize the alex variable, alex1 hasn't been set yet. You need to put initialization of alex1 first, then you'll get the expected result.
This situation is covered by JLS 8.3.2.3 "Restrictions on the use of Fields during Initialization".
The JLS rules allows the usage in your Question, and state that the first call to getc() will return default (uninitialized) value of alex.
However, the rules disallow some uses of uninitialized variables; e.g.
int i = j + 1;
int j = i + 1;
is disallowed.
Re some of the other answers. This is not a case where the Java compiler "can't figure it out". The compiler is strictly implementing what the Java Language Specification specifies. (Or to put it another way, a compiler could be written to detect the circularity in your example and call it a compilation error. However, if it did this, it would be rejecting valid Java programs, and therefore wouldn't be a conformant Java compiler.)
In a comment you state this:
... final fields always must be initialized at compile or at runtime before the object creation.
This is not correct.
There are actually two kinds of final fields:
A so-called "constant variable" is indeed evaluated at compile time. (A constant variable is a variable "of primitive type or type String, that is final and initialized with a compile-time constant expression" - see JLS 4.12.4.). Such a field will always have been initialized by the time you access it ... modulo certain complications that are not relevant here.
Other final fields are initialized in the order specified by the JLS, and it is possible to see the field's value before it has been initialized. The restriction on final variables is that they must be initialized once and only once during class initialization (for a static) or during object initialization.
Finally, this stuff is very much "corner case" behavior. A typical well-written class won't need to
access a final field before it has been initialized.
Static final fields whose values are not compile-time constant expressions are initialized in order of declaration. Thus when alex in being initialized, alex1 is not initialized yet, so that getc() returns default values of alex1 (0).
Note that result will be different (10 10 10) in the following case:
static final int alex1 = 10;
In this case alex1 is initialized by a compile-time constant expression, therefore it's initialized from the very beginning.
There is nothing special about static fields, it just that the compiler cannot workout that you are using a method which can access a field before its initialised.
e.g.
public class Main {
private final int a;
public Main() {
System.out.println("Before a=10, a="+getA());
this.a = 10;
System.out.println("After a=10, a="+getA());
}
public int getA() {
return a;
}
public static void main(String... args) {
new Main();
}
}
prints
Before a=10, a=0
After a=10, a=10
Class variables are not necessary to initialize, they are automatically set to their default values. If primitives (like int, short...) it's 0 (zero) for Objects it's null.
Therefore alex1 is set to 0.
Method variables must be initialized, otherwise you will get an compiliation error.
For a better explanation read http://download.oracle.com/javase/tutorial/java/javaOO/classvars.html