Why do fields seem to be initialized before constructor?

Question

public class Dog {

 public static Dog dog = new Dog();
 static final int val1 = -5;
 static int val2 = 3;
 public int val3;

 public Dog() {
      val3 = val1 + val2;
 }

public static void main(String[] args) {
    System.out.println(Dog.dog.val3);
}
}

The output is -5

From this result, it seems that the initialization of val2 is before the completion of dog member and its instantiating.

Why is this order like this?

Answer 1

If you move your dog instance in the end, you may find the output becomes -2

public class Dog {

     static final int val1 = -5;// This is final, so will be initialized at compile time 
     static int val2 = 3;
     public int val3;

     public static Dog dog = new Dog();//move to here
 
     public Dog() {
          val3 = val1 + val2;
     }

    public static void main(String[] args) {
        System.out.println(Dog.dog.val3);//output will be -2
    }
}

The final fields(whose values are compile-time constant expressions) will beinitialized first, and then the rest will be executed at in textual order.

So , in your case when dog instance is initialized, static int val2(0) is not initialized yet, while static final int val1(-5) does since it is final.

http://docs.oracle.com/javase/specs/jls/se5.0/html/execution.html#12.4.2 states that:

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, except that final class variables and fields of interfaces whose values are compile-time constants are initialized first

---

Updated a newer doc

Here is the jdk7 version from http://docs.oracle.com/javase/specs/jls/se7/html/jls-12.html#jls-12.4.2

final field is at step6:

Then, initialize the final class variables and fields of interfaces whose values are compile-time constant expressions

while static field is at step 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.

Answer 2

Variable declaration sequence. The static final int val1 is initialized first because it is a constant. However static int val2 is still 0 at the time public static Dog dog = new Dog(); is instantiated.

Answer 3

What is happening..

The first line that is executing is this public static Dog dog = new Dog();. Now, there are 2 things that must be kept in mind.

1. final int makes it a compile time constant. So, -5 is already hard-coded into your code.

2. The call to new Dog() is done and the constructor is called which sets the value to 0 + -5 = -5.

change val2 to final then you will see the difference (you will get -2 as answer.)

Note : static fields are initialized as and how they are encountered.