Why does a return in `finally` override `try`?

Finally always executes. That's what it's for, which means its return value gets used in your case.

You'll want to change your code so it's more like this:

function example() {     var returnState = false; // initialization value is really up to the design    try {         returnState = true;     }     catch {        returnState = false;    }    finally {         return returnState;     } } 

Generally speaking you never want to have more than one return statement in a function, things like this are why.

According to ECMA-262 (5ed, December 2009), in pp. 96:

The production TryStatement : try Block Finally is evaluated as follows:

1. Let B be the result of evaluating Block.
2. Let F be the result of evaluating Finally.
3. If F.type is normal, return B.
4. Return F.

And from pp. 36:

The Completion type is used to explain the behaviour of statements (break, continue, return and throw) that perform nonlocal transfers of control. Values of the Completion type are triples of the form (type, value, target), where type is one of normal, break, continue, return, or throw, value is any ECMAScript language value or empty, and target is any ECMAScript identifier or empty.

It's clear that return false would set completion type of finally as return, which cause try ... finally to do 4. Return F.

When you use finally, any code within that block fires before the method exits. Because you're using a return in the finally block, it calls return false and overrides the previous return true in the try block.

(Terminology might not be quite right.)