I understand the purpose of static keyword but regretfully I can't figure out this simple code. Don't know why? I'm expecting the answer to be really simple.
public class VariableScope {
int x=y;
static int y=5;
public static void main(String[] args) {
System.out.println(new VariableScope().x);
}
}
How x gets printed as 5 when y was assigned to it at an earlier stage?
The static initialization static int y = 5 happens during class loading.
The x = y assignment happens during the instance construction new VariableScope(); so at this point y has the value 5 already.
It's because order the lines of the code does not matter here. Static member is initialized first, then x is initialized.
Here is detailed specification of the initialization order: Java Language Specification 12.4.2. Detailed Initialization Procedure
Related
I would like to pass a reference to a primitive type to a method, which may change it.
Consider the following sample:
public class Main {
Integer x = new Integer(42);
Integer y = new Integer(42);
public static void main(String[] args) {
Main main = new Main();
System.out.println("x Before increment: " + main.x);
// based on some logic, call increment either on x or y
increment(main.x);
System.out.println("x after increment: " + main.x);
}
private static void increment(Integer int_ref) {
++int_ref;
}
}
The output running the sample is:
x Before increment: 42
x after increment: 42
Which means int_ref was past to the function by value, and not by reference, despite my optimistic name.
Obviously there are ways to work around this particular example, but my real application is way more complex, and in general one would imagine that a "pointer" or reference to integer would be useful in many scenarios.
I've tried to pass Object to the function (then casting to int), and various other methods, with no luck. One workaround that seems to be working would be to define my own version of Integer class:
private static class IntegerWrapper {
private int value;
IntegerWrapper(int value) { this.value = value; }
void plusplus() { ++value; }
int getValue() { return value; }
}
Doing this, and passing a reference to IntegerWrapper does work as expected, but to my taste it seems very lame. Coming from C#, where boxed variable just remain boxed, I hope I just miss something.
EDIT:
I would argue my question isn't a duplicate of Is Java "pass-by-reference" or "pass-by-value"?, as my question isn't theoretical, as I simply seek a solution. Philosophically, all method calls in all languages are pass-by-value: They either pass the actual value, or a reference to the value - by value.
So, I would rephrase my question: What is the common paradigm to workaround the issue that in java I'm unable to pass a reference to an Integer. Is the IntegerWrapper suggested above a known paradigm? Does a similar class (maybe MutableInt) already exist in the library? Maybe an array of length 1 a common practice and has some performance advantage? Am I the only person annoyed by the fact he can store a reference to any kind of object, but the basic types?
Integer is immutable, as you may notice.
Your approach with private static class IntegerWrapper is correct one. Using array with size 1 is also correct, but in practice I have never seen using array for this case. So do use IntegerWrapper.
Exactly the same implementation you can find in Apache org.apache.commons.lang3.mutable.MutableInt.
In your example you also can provide Main instance to the static method:
public class Main {
private int x = 42;
public static void main(String[] args) {
Main main = new Main();
incrementX(main);
}
private static void incrementX(Main main) {
main.x++;
}
}
And finally, from Java8 you could define an inc function and use it to increment value:
public class Main {
private static final IntFunction<Integer> INC = val -> val + 1;
private int x = 42;
public static void main(String[] args) {
Main main = new Main();
main.x = INC.apply(main.x);
}
}
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Java static class initialization
in what order are static blocks and static variables in a class executed?
When I run this code the answer is 1, I thought it would be 2.
What is the order of initialization and the value of k in each step?
public class Test {
static {k = 2;}
static int k = 1;
public static void main(String[] args) {
System.out.println(k);
}
}
Edit 1: As a follow up to "k is set to default value" then why this next code doesn't compile? Theres an error "Cannot reference a field before it's defined".
public class Test {
static {System.out.println(k);}
static int k=1;
public static void main(String[] args) {
System.out.println(k);
}
}
Edit 2: For some unknow to me reason it^ works when instead of "k" its "Test.k".
Thanks for all the answers. this will sufice :D
They are executed in the order that you write them. If the code is:
public class Test {
static int k = 1;
static {k = 2;}
public static void main(String[] args) {
System.out.println(k);
}
}
then the output becomes 2.
The order of initialization is: ..the class variable initializers and static initializers of the class..., in textual order, as though they were a single block.
And the values (for your code) are: k = 0 (default), then it's set to 2, then it's set back to 1.
You can check that it's actually set to 2 by running the following code:
private static class Test {
static {
System.out.println(Test.k);
k = 2;
System.out.println(Test.k);
}
static int k = 1;
public static void main(String[] args) {
System.out.println(k);
}
}
Short answer
When the initialization of the class starts, k will have initial value of 0.
The static block (since it precedes the assignment in the declaration) is then executed, and k will be assigned 2.
Then the initializer in the declaration is executed, and k will be assigned 1.
Long explanation
Let us use this example, since your example is a bit simple:
class TestInitOrder {
static {
System.out.println(TestInitOrder.stat1);
System.out.println(TestInitOrder.stat2);
System.out.println(TestInitOrder.str);
System.out.println(TestInitOrder.str2);
str = "something";
System.out.println(TestInitOrder.str);
System.out.println(TestInitOrder.str2);
System.out.println(TestInitOrder.lazy);
System.out.println(TestInitOrder.second);
}
private static final int stat1 = 10;
static final String str2 = "sdfff";
static String str = "crap";
private static int stat2 = 19;
static final Second second = new Second();
static final int lazy;
static {
lazy = 20;
}
static {
System.out.println(TestInitOrder.str2);
System.out.println(TestInitOrder.stat2);
System.out.println(TestInitOrder.str);
System.out.println(TestInitOrder.lazy);
System.out.println(TestInitOrder.second);
}
public static void main(String args[]) {
}
}
class Second {
public Second() {
System.out.println(TestInitOrder.second);
}
}
According to Java Language Specification, from section 4.12.5:
Every variable in a program must have a value before its value is used:
Each class variable, instance variable, or array component is initialized with a default value when it is created
(The following lines from the specification specify the default value for all the types, basically some form of 0, such as 0, 0.0d, null, false, etc.)
So before the class is initialized (due to one of these reasons), the variables will hold an initial value.
According to the detailed initialization procedure (only the interesting steps are quoted here, and emphasis mine):
6.
[...] Then, initialize the final class variables and fields of interfaces whose values are compile-time constant expressions (§8.3.2.1, §9.3.1, §13.4.9, §15.28).
[...]
9.
Next, execute either the class variable initializers and static initializers of the class, or the field initializers of the interface, in textual order, as though they were a single block.
Let us look at step 6, with the 4 final class variables: stat1, str2, second, lazy.
Since 10 is constant expression, and so is "sdfff", and due to the order of execution, it is not possible to observe the initial value for str2 and stat1. In order to make an observation, the earliest you can do is in step 9.
The case of second demonstrate that when the right hand side is not a compile-time constant expression, so its initial value is visible.
The case of lazy is different, since the assignment is done in static block, and hence happen in step 9 - so it is possible to observe its initial value. (Well, the compiler checks carefully that lazy is assigned exactly once).
After the initialization of final class variables with compile-time constant expression comes the execution of static blocks and the rest of the initializers.
As you can see from the example, the static blocks and initialization happens according to textual order - demonstrated with the use of str variable - it is first printed out as null, then something, then crap.
The question is from the famous SCJP 6 book
Given:
public class Dark {
int x = 3;
public static void main(String[] args) {
new Dark().go1();
}
void go1() {
int x;
go2(++x);
}
void go2(int y) {
int x = ++y;
System.out.println(x);
}
}
What is the result?
A. 2
B. 3
C. 4
D. 5
E. Compilation fails
F. An exception is thrown at runtime
The answer according to the book is:
✓ E is correct. In go1() the local variable x is not initialized.
My questions is why go1() cannot use instance variable x initialized as 6 on line 4 here?
Because the local variable x exists. If int x; were commented out, it will run fine and use the instance variable.
In Java all the local variables should be initialized if not it will give an error. But you shouldn't initialize parameters of a method.
if you don't have int x then this will be ok. because at that case compiler will use the local variable which is assigned for class level.
I am reading a book about Java. It gives an example like this:
class UseStatic {
static int a = 3;
static int b;
// ... some more lines
static {
b = 4 * a;
}
}
Then it explains this:
As soon as the UseStatic class is loaded, all the static
statements are run. First, a is set to 3, then the static block
executes, ..., and then initializes b to a*4 or 12.
I wonder what really happens here with the order in which the lines are excecuted. How can it first start at the top at static int a = 3 to only then jump over static int b to static {..} to only then jump back to static int b? How does it know this dependency?
It doesn't "jump over static int b", and it doesn't "jump back to static int b." There is nothing to "jump back" to because static int b is only a declaration, not an initialization/assigment. It declares the variable but does not assign it a value.
First a is declared and initialized. Then b is declared. Then the static block is executed, which assigns a value to b.
The Java Language Specification dictates the exact order of all of these things. See Jon Skeet's answer to Are there any guarantees in JLS about order of execution static initialization blocks? for more.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Why can’t enum’s constructor access static fields?
enum Test {
e1,e2;
int i=0;
static int j=5;
Test(){
System.out.println(i+" "+j);
}
}
In the above code the constructor can access the instance variable but not the static variable J.
I have read the answer relate to other author all are saying the e1 and e2 initialized before the initialization of J( static field), But according java spec all the static field
initialized when ever the class loaded to memory, that is before running of the constructor.
So before running of Test() constructor the static variable j must be initialized. I'm not able understand the restriction, can any body make me understand.I have already read the answer of the questions Why can't enum's constructor access static fields? But I am not happy with answer like :-The constructor is called before the static fields have all been initialized.
Suppose if take another example with a simple class like enum
class Test{
public static final Test t=new Test();
static int a=5;
Test(){
System.out.println(a);
}
public static void main(String[] args) {
}
}
Here according to there argument the constructor will run before the initialization of static field and it's running also as it's print 0(As JVM did the initilization). But no compilation error or no run time problem. Then why the same thing not happen with enum.
If you imagine how your enum would actually look as a class, it makes sense:
public class Test {
// Imagine you cannot move these two statements:
public static final Test e1 = new Test();
public static final Test e2 = new Test();
int i=0;
static int j=5;
private Test(){
System.out.println(i+ " " + j);
}
static int getJ() {
return j;
}
public static void main(String[] args) {
System.out.println(Test.getJ());
}
}
This prints:
0 0
0 0
5
If you can share a concrete example (rather than a theoretical one), we could suggest how to redesign the code to achieve the desired result, despite the static field limitations.
Problem is, that instances of enum are created during inicialization of static fields. And they created before initialization of your static fields.
They must be in static array values and statically accessible, so it makes sense. And as stated in anser to "Why can't enum's constructor access static fields?", its unfortunate that this happens before all user defined static field inicialization.
But if it was swapped, you could not acces enum instances in static initialization, so it would need allowing static block both before and after creation of enum values.
I don't know whether problem is because inicialization of enum values is concern of Enum class (and handled specialy by JVM (this logic is not in Enum class itself), or because you cannot put static fields before enum values.
WHY it is that way can answer only few people (eg. Josh Bloch and Neal Gafter who are stated as authors of Enum in javadoc, and maybe some unknown others)