Why can I anonymously subclass an enum but not a final class?

Question

This code:

public class Sandbox {
    public enum E {
        VALUE {
            @Override
            public String toString() {
                return "I'm the value";
            }
        };

        @Override
        public String toString() {
            return "I'm the enum";
        }
    }
    public static void main(String[] args) {
        System.out.println(E.VALUE);
    }
}

prints:

I'm the value

However, this code:

public class Sandbox {
    public static final class C {
        @Override
        public String toString() {
            return "I'm a C";
        }
    }

    public static void main(String[] args) {
        System.out.println(new C() {
            @Override
            public String toString() {
                return "I'm anonymous";
            }
        });
    }
}

results in a compilation error:

cannot inherit from final HelloWorld.C

Why can E.VALUE create what appears to me to be an anonymous E subclass, overriding the toString method, while using a final class instead of an implicitly final enum throws a compile-time error?

More specifically, why can VALUE override anything in E? I was under the impression that the code

public enum E {
    VALUE;
}

was roughly equivalent to

public static final class E {
    public static final E VALUE = new E();
}

in which case the anonymous nature would not be allowed.

What's the difference? Why are enums special?

Answer 1

According to the JLS:

An enum type is implicitly final unless it contains at least one enum constant that has a class body.

In your example, VALUE has a class body and therefore E is not implicitly final.

Edit: Here's a quick example that validates the claim:

import java.lang.reflect.Modifier;

public class Sandbox {
  public enum E {
    VALUE {};
  }

  public enum E2 {
    VALUE;
  }

  public static void main(String[] args) {
    System.out.println(E.class);
    System.out.println(E.VALUE.getClass());
    System.out.println("E.VALUE is subclass of E = " + E.VALUE.getClass().getSuperclass().equals(E.class));
    System.out.println("E modifiers: " + Modifier.toString(E.class.getModifiers()));
    System.out.println("E2 modifiers: " + Modifier.toString(E2.class.getModifiers()));
  }
}

You can see from the output that the compiler is adding the final modifier to E2 but not to E:

class Sandbox$E
class Sandbox$E$1
E.VALUE is subclass of E = true
E modifiers: public static
E2 modifiers: public static final

Edit #2: Even though E is not final and is subclassed by VALUE, explicitly trying to extend it such as with class Foo extends E or enum Bar extends E is a compile-time error according to 8.1.4. Superclasses and Subclasses:

It is a compile-time error if the ClassType names the class Enum or any invocation of it.

Answer 2

I think if you compare classes and enums, then the enum E can be compared to a class, and the enum value VALUE can be compared to an anonymous instance. Thus, your first example can be rewritten as following:

public class Sandbox {
    public static class E {
        public static final E VALUE = new E() {
            @Override
            public String toString() {
                return "I'm the value";
            }
        };

        @Override
        public String toString() {
           return "I'm the enum";
       }
    }
    public static void main(String[] args) {
        System.out.println(E.VALUE);
    }
}

Answer 3

If you take the following into account, your concern may be lessened:

A class which has no non-private constructor is effectively final since it is impossible to declare its subclass outside of its body. It is a minor technical detail whether the class defined by an enum is declared final or not: it will in either case be effectively final.