java - 为什么字段似乎在构造函数之前被初始化？

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);
}
}

输出是-5

从这个结果看来，val2 的初始化似乎是在 dog 成员及其实例化完成之前。

为什么这个顺序是这样的？

Answer 1

如果你最后移动你的狗实例，你可能会发现输出变成-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
    }
}

最终的字段(其值为编译时常量表达式)将首先被初始化，然后其余的将按文本顺序执行。

因此，在您初始化狗实例的情况下， static int val2(0) 尚未初始化，而 static final int val1(-5) 则自这是最终的。

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

execute either the class variable initializers and staticinitializers of the class, or the field initializers of the interface,in textual order, as though they were a single block, except thatfinal class variables and fields of interfaces whose values arecompile-time constants are initialized first

---

更新了一个较新的文档

这里是 http://docs.oracle.com/javase/specs/jls/se7/html/jls-12.html#jls-12.4.2 中的 jdk7 版本

final 字段在第 6 步:

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

当静态字段在第 9 步时:

Next, execute either the class variable initializers and staticinitializers of the class, or the field initializers of the interface,in textual order, as though they were a single block.