How does a return statement inside a try/catch block work?
function example() {
try {
return true;
}
finally {
return false;
}
}
I'm expecting the output of this function to be true
, but instead it is false
!
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:
- Let B be the result of evaluating Block.
- Let F be the result of evaluating Finally.
- If F.type is normal, return B.
- Return F.
And from pp. 36:
The Completion type is used to explain the behaviour of statements (
break
,continue
,return
andthrow
) that perform nonlocal transfers of control. Values of the Completion type are triples of the form (type, value, target), where type is one ofnormal
,break
,continue
,return
, orthrow
, 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.
The finally block rewrites try block return (figuratively speaking).
Just wanted to point out, that if you return something from finally, then it will be returned from the function. But if in finally there is no 'return' word - it will be returned the value from try block;
function example() {
try {
return true;
}
finally {
console.log('finally')
}
}
console.log(example());
// -> finally
// -> true
So -finally- return
rewrites the return of -try- return
.
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.
Returning from a finally-block
If the
finally
-block returns a value, this value becomes the return value of the entiretry-catch-finally
statement, regardless of anyreturn
statements in thetry
andcatch
-blocks
Reference: developer.mozilla.org
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:
res.send()
from Express.js, which creates a HTTP response and dispatches it.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
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.