How final variables in method can be accessed by Inner class even if method goes out of scope?

Question

I have already gone through many links for this answer. And I know that syntactically Java does not allow non-final variables to be accessed in Inner classes. According to this link, I have also understood that making a local-variable final allows it to stay alive even if the method gets completed and local variable goes out of scope.

But if a method goes out of scope how making it final would allow it to be alive for Inner classes where it has been referenced?

I am referring to the below code.

public class Outer {
    public static void outerMethod() {
        int methodVariable = 40;
        class Inner {
            public void innerMethod() {
                System.out.println(methodVariable); // This line produces compile time error.
            }
        }
    }
}

The code in the question actually compiles now, although this has no bearing on your question of how the inner class is able to access the variable after the method returns. — Radiodef, Mar 17 '18 at 17:30

score 2 · Accepted Answer · answered Mar 17 '18 at 17:35

If you check an instance of the Inner class in a debugger, you'll spot that there is a field for methodVariable:

Not that it is the only possible implementation, but it seems the compiler actually creates a private field for methodVariable and copies the value of the variable there.

This is probably also why the variable must be final or effectively final. If the value of the variable changes, this can't longer be synchronized with the value of the field.

How final variables in method can be accessed by Inner class even if method goes out of scope?

1 Answers1