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

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栀梦
栀梦 2020-12-12 15:22

How does a return statement inside a try/catch block work?

function example() {
    try {
        return true;
    }
    finally {
        return false;
             


        
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  • 2020-12-12 16:09

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

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

    function example() { 
        var returnState = false; // initialisation 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.

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  • 2020-12-12 16:09

    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.

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  • 2020-12-12 16:13

    As far as I know, the finally block always executes, irrespective of whether you have a return statement inside try or not. Ergo, you get the value returned by the return statement inside finally block.

    I tested this with Firefox 3.6.10 and Chrome 6.0.472.63 both in Ubuntu. It is possible that this code may behave differently in other browsers.

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  • 2020-12-12 16:16

    I'm gonna give a slightly different answer here: Yes, both the try and finally block get executed, and finally takes precedence over the actual "return" value for a function. However, these return values aren't always used in your code.

    Here's why:

    • The example below will use res.send() from Express.js, which creates a HTTP response and dispatches it.
    • Your try and finally block will both execute this function like so:
    try {
        // Get DB records etc.
        return res.send('try');
    } catch(e) {
        // log errors
    } finally {
        return res.send('finally');
    }
    

    This code will show the string try in your browser. ALSO, the example will show an error in your console. The res.send() function is called twice. This will happen with anything that is a function. The try-catch-finally block will obfuscate this fact to the untrained eye, because (personally) I only associate return values with function-scopes.

    Imho your best bet is to never use return inside a finally block. It will overcomplicate your code and potentially mask errors.

    In fact, there's a default code inspection rule set-up in PHPStorm that gives a "Warning" for this:

    https://www.jetbrains.com/help/phpstorm/javascript-and-typescript-return-inside-finally-block.html

    So what do you use finally for?

    I would use finally only to clean-up stuff. Anything that is not critical for the return value of a function.

    It may make sense if you think about it, because when you depend on a line of code under finally, you are assuming that there could be errors in try or catch. But those last 2 are the actual building blocks of error handling. Just use a return in try and catch instead.

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