|
|
|
|
|
by desert_boi
2419 days ago
|
|
Interestingly when I was encountering this myself recently, I discovered that JS finally blocks can return after a function has nominally already returned. Consider the following closure. (() => {
try {
// finally will return prior to this console.log
console.log('this try was executed and the return ignored')
return 'try block'
} catch (e) {
return 'error block'
} finally {
return 'finally block'
}
})()
|
|
Basically, `finally` gives you a guarantee that it will actually run once the `try` block is exited. Likewise, `return` effectively assigns the return value and exits. But it doesn't (cannot and should not) breach the contract of try-finally, since the purpose of try-finally is to ensure resources are managed correctly, so it either exits to the caller or it exits to the finally block, depending on which one is most recent.
In your case, a return value is assigned and the `try` block is exited using `return`. We then have to continue into the `finally` block, since that is the core meaning of `finally` - we run it after we leave `try`. And then with `return`, we reassign the return value and finally leave the whole function. At this point, the return value is the second one that was assigned to it.
Maybe thinking of it like this is helpful, although I somewhat hope it isn't. You can see that "return" is reassigned before we have a chance to read it. I've simplified by removing any consideration of errors, but I console.logged the final output.