Why is a local variable in Java not considered "effectively final" even though nothing modifies it afterwards?

Question

In a method I have this:

int x = 0
if (isA()) {
    x = 1;
} else if (isB()) {
    x = 2;
}

if (x != 0) {
    doLater(() -> showErrorMessage(x)); // compile error here
}

// no more reference to 'x' here

I don't understand why it produces compilation error. The error says that x is not final or effectively-final, so it can't be accessed from the lambda body. There is no modification to x after the doLater call, so the value of x is actually already determined when doLater is called.

I am guessing that the answer to this question is because x is not qualified to be called an effectively-final variable. However, I want to know what the reason is.

Can't the compiler just create a temporary final variable, effectively making the code like:

if (x != 0) {
    final int final_x = x;
    doLater(() -> showErrorMessage(final_x));
}

and everything still work the same?

Answer 1

Effectively final means that it could have been made final i.e. it never changes. It means that effectively, the variable could be a final one.

The problem is that it doesn't keep track of the last time you changed it, but rather, did you ever change it. Change your if statement to

int x;
if (isA()) {
    x = 1;
} else if (isB()) {
    x = 2;
}  else {
    x = 0;
}

or

int x = isA() ? 1 : 
        isB() ? 2 : 0;

Answer 2

Your x variable would have been effectively final it it was initialized once and not changed again under any circumstances. If you had only:

int x = 0;
doLater(() -> showErrorMessage(x));

then it would have compiled.

However, adding conditions that might change the variable's value

int x = 0;
if (isA()) {
    x = 1;
} else if (isB()) {
    x = 2;
}

makes the variable being not effectively final and thus the compile error is risen.

---

Additionally, since this pointer approach you've implemented wouldn't work, you could refactor your code a bit to a simple if-else statement:

if (isA()) {
    doLater(() -> showErrorMessage(1));
} else if (isB()) {
    doLater(() -> showErrorMessage(2));
}

and completely get rid of x.

Answer 3

Short version, a variable is effectively final if it is assigned exactly once, no matter which code path is executed.

Long version, quoting Java Language Specification 4.12.4. final Variables (emphasis mine):

Certain variables that are not declared final are instead considered effectively final:

• A local variable whose declarator has an initializer (§14.4.2) is effectively final if all of the following are true:
  • It is not declared final.
  • It never occurs as the left hand side in an assignment expression (§15.26). (Note that the local variable declarator containing the initializer is not an assignment expression.)
  • It never occurs as the operand of a prefix or postfix increment or decrement operator (§15.14, §15.15).

Now, you can make it effectively final by removing the initializer, because it continues:

• A local variable whose declarator lacks an initializer is effectively final if all of the following are true:
  • It is not declared final.
  • Whenever it occurs as the left hand side in an assignment expression, it is definitely unassigned and not definitely assigned before the assignment; that is, it is definitely unassigned and not definitely assigned after the right hand side of the assignment expression (§16 (Definite Assignment)).
  • It never occurs as the operand of a prefix or postfix increment or decrement operator.