Understanding enum's static members initialization

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Execution order of of static blocks in an Enum type wrt to constructor 3 answers

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The enum's constants are themselves initialized by calling the constructor. This means the constructor cannot access the constant since it's not yet created at that time.

In other words, say you have:

enum MyEnum {

   FOO, BAR;

   private MyEnum() {
        // Illegal
        // FOO already calls this constructor
        System.out.println(FOO);
   }

}

FOO and BAR are equivalent to:

public static final MyEnum foo;
public static final MyEnum bar;

When the enum class is loaded by the JVM, FOO and BAR are created by calling the enum private constructor, something like:

foo = MyEnum(); // name of enum, the params are not relevant
bar = MyEnum(); 

So Java does not allow you to access that field in the constructor since it is still under creation. You can run the following to verify:

enum MyEnum {
    FOO, BAR;

    private MyEnum() {
        System.out.println("Initializing");
    }
}

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

Output:

Initializing
Initializing
FOO

"Initializing" is printed twice, one by the creation of FOO and one by BAR .

The JLS also says this about enums:

It is a compile-time error to reference a static field of an enum type from constructors, instance initializers, or instance variable initializer expressions of the enum type, unless the field is a constant variable (§4.12.4).

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