Is final ill-defined?

Question

First, a puzzle: What does the following code print?

public class RecursiveStatic {
    public static void main(String[] args) {
        System.out.println(scale(5));
    }

    private static final long X = scale(10);

    private static long scale(long value) {
        return X * value;
    }
}

Answer:

0

Spoilers below.

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If you print X in scale(long) and redefine X = scale(10) + 3, the prints will be X = 0 then X = 3. This means that X is temporarily set to 0 and later set to 3. This is a violation of final!

The static modifier, in combination with the final modifier, is also used to define constants. The final modifier indicates that the value of this field cannot change.

Source: https://docs.oracle.com/javase/tutorial/java/javaOO/classvars.html [emphasis added]

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My question: Is this a bug? Is final ill-defined?

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Here is the code that I am interested in. X is assigned two different values: 0 and 3. I believe this to be a violation of final.

public class RecursiveStatic {
    public static void main(String[] args) {
        System.out.println(scale(5));
    }

    private static final long X = scale(10) + 3;

    private static long scale(long value) {
        System.out.println("X = " + X);
        return X * value;
    }
}

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This question has been flagged as a possible duplicate of Java static final field initialization order. I believe that this question is not a duplicate since the other question addresses the order of initialization while my question addresses a cyclic initialization combined with the final tag. From the other question alone, I would not be able to understand why the code in my question does not make an error.

This is especially clear by looking at the output that ernesto gets: when a is tagged with final, he gets the following output:

a=5
a=5

which does not involve the main part of my question: How does a final variable change its variable?

Answer 1

Reading an uninitialized field of an object ought to result in a compilation error. Unfortunately for Java, it does not.

I think the fundamental reason why this is the case is "hidden" deep within the definition of how objects are instantiated and constructed, though I don't know the details of the standard.

In a sense, final is ill-defined because it doesn't even accomplish what its stated purpose is due to this problem. However, if all your classes are properly written, you don't have this problem. Meaning all fields are always set in all constructors and no object is ever created without calling one of its constructors. That seems natural until you have to use a serialization library.

Answer 2

Class level members can be initialized in code within the class definition. The compiled bytecode cannot initialize the class members inline. (Instance members are handled similarly, but this is not relevant for the question provided.)

When one writes something like the following:

public class Demo1 {
    private static final long DemoLong1 = 1000;
}

The bytecode generated would be similar to the following:

public class Demo2 {
    private static final long DemoLong2;

    static {
        DemoLong2 = 1000;
    }
}

The initialization code is placed within a static initializer which is run when the class loader first loads the class. With this knowledge, your original sample would be similar to the following:

public class RecursiveStatic {
    private static final long X;

    private static long scale(long value) {
        return X * value;
    }

    static {
        X = scale(10);
    }

    public static void main(String[] args) {
        System.out.println(scale(5));
    }
}

1. The JVM loads the RecursiveStatic as the jar's entry point.
2. The class loader runs the static initializer when the class definition is loaded.
3. The initializer calls the function scale(10) to assign the static final field X.
4. The scale(long) function runs while the class is partially initialized reading the uninitialized value of X which is the default of long or 0.
5. The value of 0 * 10 is assigned to X and the class loader completes.
6. The JVM runs the public static void main method calling scale(5) which multiplies 5 by the now initialized X value of 0 returning 0.

The static final field X is only assigned once, preserving the guarantee held by the final keyword. For the subsequent query of adding 3 in the assignment, step 5 above becomes the evaluation of 0 * 10 + 3 which is the value 3 and the main method will print the result of 3 * 5 which is the value 15.

Answer 3

A very interesting find. To understand it we need to dig into the Java Language Specification (JLS).

The reason is that final only allows one assignment. The default value, however, is no assignment. In fact, every such variable (class variable, instance variable, array component) points to its default value from the beginning, before assignments. The first assignment then changes the reference.

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Class variables and default value

Take a look at the following example:

private static Object x;

public static void main(String[] args) {
    System.out.println(x); // Prints 'null'
}

We did not explicitly assign a value to x, though it points to null, it's default value. Compare that to §4.12.5:

Initial Values of Variables

Each class variable, instance variable, or array component is initialized with a default value when it is created (§15.9, §15.10.2)

Note that this only holds for those kind of variables, like in our example. It does not hold for local variables, see the following example:

public static void main(String[] args) {
    Object x;
    System.out.println(x);
    // Compile-time error:
    // variable x might not have been initialized
}

From the same JLS paragraph:

A local variable (§14.4, §14.14) must be explicitly given a value before it is used, by either initialization (§14.4) or assignment (§15.26), in a way that can be verified using the rules for definite assignment (§16 (Definite Assignment)).

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Final variables

Now we take a look at final, from §4.12.4:

final Variables

A variable can be declared final. A final variable may only be assigned to once. It is a compile-time error if a final variable is assigned to unless it is definitely unassigned immediately prior to the assignment (§16 (Definite Assignment)).

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Explanation

Now coming back to the your example, slightly modified:

public static void main(String[] args) {
    System.out.println("After: " + X);
}

private static final long X = assign();

private static long assign() {
    // Access the value before first assignment
    System.out.println("Before: " + X);

    return X + 1;
}

It outputs

Before: 0
After: 1

Recall what we have learned. Inside the method assign the variable X was not assigned a value to yet. Therefore, it points to its default value since it is an class variable and according to the JLS those variables always immediately point to their default values (in contrast to local variables). After the assign method the variable X is assigned the value 1 and because of final we can't change it anymore. So the following would not work due to final:

private static long assign() {
    // Assign X
    X = 1;

    // Second assign after method will crash
    return X + 1;
}

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Example in the JLS

Thanks to @Andrew I found a JLS paragraph that covers exactly this scenario, it also demonstrates it.

But first let's take a look at

private static final long X = X + 1;
// Compile-time error:
// self-reference in initializer

Why is this not allowed, whereas the access from the method is? Take a look at §8.3.3 which talks about when accesses to fields are restricted if the field was not initialized yet.

It lists some rules relevant for class variables:

For a reference by simple name to a class variable f declared in class or interface C, it is a compile-time error if:

• The reference appears either in a class variable initializer of C or in a static initializer of C (§8.7); and

• The reference appears either in the initializer of f's own declarator or at a point to the left of f's declarator; and

• The reference is not on the left hand side of an assignment expression (§15.26); and

• The innermost class or interface enclosing the reference is C.

It's simple, the X = X + 1 is caught by those rules, the method access not. They even list this scenario and give an example:

Accesses by methods are not checked in this way, so:

class Z {
    static int peek() { return j; }
    static int i = peek();
    static int j = 1;
}
class Test {
    public static void main(String[] args) {
        System.out.println(Z.i);
    }
}

produces the output:

0

because the variable initializer for i uses the class method peek to access the value of the variable j before j has been initialized by its variable initializer, at which point it still has its default value (§4.12.5).

Answer 4

Nothing to do with final here.

Since it is at instance or class level, it holds the default value if nothing gets assigned yet. That is the reason you seeing 0 when you accessing it without assigning.

If you access X without completely assigning, it holds the default values of long which is 0, hence the results.

Answer 5

It's not a bug at all, simply put it is not an illegal form of forward references, nothing more.

String x = y;
String y = "a"; // this will not compile 


String x = getIt(); // this will compile, but will be null
String y = "a";

public String getIt(){
    return y;
}

It's simply allowed by the Specification.

To take your example, this is exactly where this matches:

private static final long X = scale(10) + 3;

You are doing a forward reference to scale that is not illegal in any way as said before, but allows you to get the default value of X. again, this is allowed by the Spec (to be more exact it is not prohibited), so it works just fine

Answer 6

Not a bug.

When the first call to scale is called from

private static final long X = scale(10);

It tries to evaluate return X * value. X has not been assigned a value yet and therefore the default value for a long is used (which is 0).

So that line of code evaluates to X * 10 i.e. 0 * 10 which is 0.