try, catch, finally

[Tomas Neme]

I read this in the docs:

With this approach, you also get implicit error propagation, just like
try, catch, and finally. An error in step1 will flow all the way to
step5, where it’s caught and handled.

and I thought I could do this with a semaphore:

sem.take(function() { callPromise().fin(sem.leave) })

I was expecting that to have the semaphore be released whether there's
an error or not, but to have any unhandled errors thrown anyways, like
if I'd done

try {
} finally {
sem.leave()
}

but while I have the semaphore released, callPromise is failing silently.

How should I have this to release the semaphore regardless of errors,
but still raise any errors?

thanks
tomas

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[Domenic Denicola]

Could you give an example that doesn't involve other libraries, but instead just Q? It would be much easier to diagnose what's going wrong in that case. I'm not familiar with what terms like "sem.take" "semaphore be released", "sem.leave", etc. mean, or how they fit into this sem library you're using, and it would be nicer if we could guarantee it wasn't an issue with the interaction of this library and Q but instead with Q itself.

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[Kris Kowal]

What is likely happening is that an error is getting swallowed. Try:

sem.take(function() {

    callPromise().fin(sem.leave).done()

})

Also, you’ll find that the require("q/queue") module gives you an infinite promise queue, which could be used as a semaphore.

var Queue = require("q/queue");

var semaphore = Queue();

semaphore.get()

.then(callPromise)

.finally(semaphore.put)

.done();

Kris Kowal

[Tomas Neme]

> Also, you’ll find that the require("q/queue") module gives you an infinite
> promise queue, which could be used as a semaphore.
>
> var Queue = require("q/queue");
> var semaphore = Queue();
>
> semaphore.get()
> .then(callPromise)
> .finally(semaphore.put)
> .done();

wow, nice to know.

as for the error I was having, I just figured it was being handled in
ANOTHER .fin() further down the promise chain, sorry