Here's a good stumper. Any ideas? Java BUG?

Question

public class Test {
    //A static value we will set later -- this is legal
    static final int INT_VALUE;
    static {
        //try to set the value..
        //note that an exception *could* be thrown BEFORE the value is set
        try {
            if (getValue()>50)
                INT_VALUE = 50;
            else
                INT_VALUE = 5;
        } catch (Exception ex) {
            //if there was an exception, the value wasn't set, so we do it now
            INT_VALUE = 1; // <-- javac will complain about this. why?
    static int getValue() throws Exception {
        throw new Exception("whoops");
}If you compile this, javac tells you the line
INT_VALUE = 1; is bad because the valriable might have been assigned to.
What order of operations could throw the exception and assign the value before the
INT_VALUE = 1; line is reached?

Answer 1

If you compile this, javac tells you the line
INT_VALUE = 1; is bad because the
valriable might have been assigned to.
What order of operations could throw the exception
and assign the value before the
INT_VALUE = 1; line is reached?The person who said the java compiler is dumb is just about right. The complaint that the compiler gives is that the value might be assigned already. This is beacuse the INT_VALUE varible has a chance to be assigned in the try block.
In your code you force a deterministic error which is usually not the case so the compiler doesn't search for it. Hence the compiler says that there is a chance that the value ha already been assigned. This is also because the value assignement, if it happened, will stay past the try block and the catch will try to reassign the final.
So the compiler just plays it safe and says the variable might have already been assigned rather than searching for possible exceptions and determining if the variable gets set to a value or not (remember your code is exceptionally deterministic with respect to the runtime error being thrown.)