Does promise need call stack to be empty to get resolved - javascript

I used process.binding('util').getPromiseDetails as shown below to get promise status and I used node dns module functions that returns promise. I then queried whether that promise is resolved in synchronous manner, but result always showed promise is not resolved even though answer usually comes in milliseconds.
const { Resolver } = require('dns').promises;
const resolver = new Resolver();
let a = resolver.resolve4('example.org');
let time = Date.now();
while (time + 10000 > Date.now()) {
console.log(process.binding('util').getPromiseDetails(a));
}
My question was do we need stack to be empty so that even internally promise is marked as resolved. I know that then handler runs when stack is empty, but does internally promise is marked as resolved only when stack is empty?
If resolution of promise needs stack to be empty to get marked as resolved when stack is empty, how am I able to add a debugger; statement and wait and when I inspect promise (after waiting for a while with code still paused) it is resolved, even though stack is not empty as code is paused.
For example:
let a = fetch('google.com');
debugger;
console.log("hello");
I can now wait and see that a is fulfilled even though I have not resumed the code, so if I have not resumed the code and stack remains full, how is a then getting fulfilled?

Do we need stack to be empty so that even internally promise is marked as resolved?
No, the runtime can mark promises that it created internally as resolved at any time (though it probably needs some kind of semaphore lock on the job queue), as seen with fetch in your example.
However, the dns/promises resolve method does not construct an internal promise. It is created from JS code and the resolved from an event handler that is called scheduled from the native code. This handler code does indeed need an empty stack to run, so it has not run by the time when your code was blocked for one second and inspects the promise.

Related

What happens to callBack passed to .then()? [duplicate]

I observe deferring of completeness notifications in Firefox's promises. Following assertion fails, because onFullfilled() is called too late*.
var resolved = false;
function onFullfilled() {
resolved = true;
log("Completed");
}
Promise.resolve(true).then(onFullfilled);
assert(resolved, "Promise completed promise should call resolution hook immediately.");
When exactly onFullfilled() is guaranteed to be called on Promise resolution?
* In my case "Completed" log message appears after test framework report assertion failure.
Promise resolution hooks are always called after all sync code is executed. This is by design - and is done in order to prevent race conditions.
Since promises sometimes resolve asynchronously the specification mandates they are always resolved asynchronously so the same code path executes. Promises guard you against Zalgo.
It is specified here:
onFulfilled or onRejected must not be called until the execution context stack contains only platform code.
A lot of test frameworks - namely Mocha support testing promises directly with promise-syntax - by returning a promise.
it("does something", function(){
return aPromise; // if aPromise resolves the test passes
})
You should always give a function to "then". So you should use "onFullfilled()" instead of "onFullfilled" as parameter of "then".
So it should be like:
Promise.resolve(true).then(onFullfilled());

Working of call stack when async/await is used

How does the Call Stack behave when async/await functions are used ?
function resolveAfter2Seconds() { // taken from https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Statements/async_function
return new Promise(resolve => {
setTimeout(() => {
resolve('resolved');
}, 2000);
});
}
const asyncFuntion=async()=>{
const result = await resolveAfter2Seconds();
console.info("asyncFuntion finish, result is: ", result);
}
const first = async()=>{
await asyncFuntion();
console.log('first completed');
debugger;
}
const second = ()=>{
console.log('second completed');
debugger;
}
function main(){
first();
second();
}
main();
In the above code, when the first breakpoint is encountered in second(), I could see that the call stack contained main() and second(). And during the second breakpoint in first(), the call stack contained main() and first().
What happened to first() during the first breakpoint. Where is it pushed ? Assuming that the asyncFunction() takes some time to complete.
Someone please help.
First off, when you get to the breakpoint you hit in second, first has already executed and is no longer on the stack.
When we go into first, we instantly hit an await asyncFunction(). This tells JavaScript to not block on the result of the call, but feel free to go looking for something else to do while we're waiting. What does Javascript do?
Well, first of all, it does call asyncFunction(). This returns a promise, and will have started some asynchronous process that will later resolve it. Now we can't continue with the next line of first (ie, the console.log('first completed')) because our await means we can't carry on until the promise was fulfilled, so we need to suspend execution here and go find something else to do with our free time.
So, we look up the stack. We're still in the first() call from main, and now we can just return a promise from that call, which we will resolve when the asynchronous execution completes. main ignores that promise return value, so we continue right with second(). Once we've executed second, we look back to whatever called that, and carry on with synchronous execution in the way everyone would expect.
Then, at some point in the future, our promise fulfills. Maybe it was waiting on an API call to return. Maybe it was waiting on the database to reply back to it. In our example, it was waiting for a 2s timeout. Whatever it was waiting for, it now is ready to be dealt with, and we can un-suspend the first call and continue executing there. It's not "called" from main - internally the function is picked back up, just like a callback, but crucially is called with a completely new stack, which will be destroyed once we're done calling the remainder of the function.
Given that we're in a new stack, and have long since left the 'main' stack frame, how do main and first end up on the stack again when we hit the breakpoint inside it?
For a long time, if you ran your code inside debuggers, the simple answer was that they wouldn't. You'd just get the function you were in, and the debugger would tell you it had been called from "asynchronous code", or something similar.
However, nowadays some debuggers can follow awaited code back to the promise that it is resolving (remember, await and async are mostly just syntactic sugar on top of promises). In other words, when your awaited code finishes, and the "promise" under the hood resolves, your debugger helpfully figures out what the stack "should" look like. What it shows doesn't actually bear much resemblance to how the engine ended up calling the function - after all, it was called out of the event loop. However, I think it's a helpful addition, enabling us all to keep the mental model of our code much simpler than what's actually going on!
Some further reading on how this works, which covers much more detail than I can here:
Zero-cost async stack traces
Asynchronous stack traces

Promises not resolving in the way I would expect

I have the following:
for (let job of jobs) {
resets.push(
new Promise((resolve, reject) => {
let oldRef = job.ref
this._sequenceService.attachRef(job).then(() => {
this._dbService.saveDoc('job', job).then(jobRes => {
console.log('[%s / %s]: %s', oldRef, jobRes['jobs'][0].ref, this._moment.unix(job.created).format('DD/MM/YYYY HH:mm'))
resolve()
}).catch(error => {
reject(error)
})
}).catch(error => {
reject(error)
})
})
)
}
return Promise.all(resets).then(() => {
console.log('Done')
})
this._sequenceService.attachRef has a console.log() call.
When this runs, I am seeing all the console logs from the this._sequenceService.attachRef() call and then I see all the logs in the saveDoc.then() call. I was expecting to see them alternate. I understand that according to this article, promises don't resolve in order but I wouldn't expect my promise to resolve until I call resolve() so would still expect alternating logs, even if not in order.
Why is this not the case?
Your code can be written a lot cleaner by avoiding the promise anti-pattern of wrapping a promise in a new manually created promise. Instead, you just push the outer promise into your array and chain the inner promises to the outer one by returning them from inside the .then() handler. It can all be done as simply as this:
for (let job of jobs) {
let oldRef = job.ref;
resets.push(this._sequenceService.attachRef(job).then(() => {
// chain this promise to parent promise by returning it
// inside the .then() handler
return this._dbService.saveDoc('job', job).then(jobRes => {
console.log('[%s / %s]: %s', oldRef, jobRes['jobs'][0].ref, this._moment.unix(job.created).format('DD/MM/YYYY HH:mm'));
});
}));
}
return Promise.all(resets).then(() => {
console.log('Done')
}).catch(err => {
console.log(err);
});
Rejections will automatically propagate upwards so you don't need any .catch() handlers inside the loop.
As for sequencing, here's what happens:
The for loop is synchronous. It just immediately runs to completion.
All the calls to .attachRef() are asynchronous. That means that calling them just initiates the operation and then they return and the rest of your code continues running. This is also called non-blocking.
All .then() handlers are asynchronous. The earliest they can run is on the next tick.
As such this explains why the first thing that happens is all calls to .attachRef() execute since that's what the loop does. It immediately just calls all the .attachRef() methods. Since they just start their operation and then immediately return, the for loop finishes it's work pretty quickly of launching all the .attachRef() operations.
Then, as each .attachRef() finishes, it will trigger the corresponding .saveDoc() to be called.
The relative timing between when .saveDoc() calls finish is just a race depending upon when they got started (how long their .attachRef() that came before them took) and how long their own .saveDoc() call took to execute. This relative timing of all these is likely not entirely predictable, particularly if there's a multi-threaded database behind the scenes that can process multiple requests at the same time.
The fact that the relative timing is not predictable should not be a surprise. You're running multiple two-stage async operations purposely in parallel which means you don't care what order they run or complete in. They are all racing each other. If they all took exactly the same time to execute with no variation, then they'd probably finish in the same order they were started, but any slight variation in execution time could certainly change the completion order. If the underlying DB gets has lock contention between all the different requests in flight at the same time, that can drastically change the timing too.
So, this code is designed to do things in parallel and notify you when they are all done. Somewhat by definition, that means that you aren't concerned about controlling the precise order things run or complete in, only when they are all done.
Clarifying my comments above:
In this case the promises are at four levels.
Level 0: Promise created with Promise.all
Level 1: Promise created with new Promise for each Job
Level 2: Promise as generated by this._sequenceService.attachRef(job)
Level 3: Promise as generated by this._dbService.saveDoc('job', job)
Let's say we have two jobs J1 and J2
One possible order of execution:
L0 invoked
J1-L1 invoked
J2-L1 invoked
J1-L2 invoked
J2-L2 invoked
J1-L2 resolves, log seen at L2 for J1
J1-L3 invoked
J2-L2 resolves, log seen at L2 for J2
J2-L3 invoked
J1-L3 resolves, log seen at L3 for J1
J1-L1 resolves
J2-L3 resolves, log seen at L3 for J2
J2-L1 resolves
L0 resolves, 'Done' is logged
Which is probably why you see all L2 logs, then all L3 logs and then finally the Promise.all log

Rejected promises in Protractor/WebDriverJS

WebDriverJS and Protractor itself are entirely based on the concept of promises:
WebDriverJS (and thus, Protractor) APIs are entirely asynchronous. All
functions return promises.
WebDriverJS maintains a queue of pending promises, called the control
flow, to keep execution organized.
And, according to the definition:
A promise is an object that represents a value, or the eventual
computation of a value. Every promise starts in a pending state and
may either be successfully resolved with a value or it may be rejected
to designate an error.
The last part about the promise rejection is something I don't entirely understand and haven't dealt with in Protractor. A common pattern we've seen and written is using then() and providing a function for a successfully resolved promise:
element(by.css("#myid")).getAttribute("value").then(function (value) {
// do smth with the value
});
The Question:
Is it possible that a promise returned by any of the Protractor/WebDriverJS functions would not be successfully resolved and would be rejected? Should we actually worry about it and handle it?
I've experienced a use-case of promise rejection while using browser.wait(). Here is an example:
var EC = protractor.ExpectedConditions;
function isElementVisible() {
var el = element(by.css('#myel'));
// return promise
return browser.wait(EC.visibilityOf(el), 1000)
.then(function success() {
return true; // return if promise resolved
}, function fail() {
return false; // return if promise rejected
});
}
expect(isElementVisible()).toBe(true);
expect(isElementVisible()).toBe(false);
Here, if element is on a page, success will be executed, otherwise, if it is not found when 1 second passes, then fail will be called. My first point is that providing a callback for rejection gives an ability to be consistent with what one should expect. In this case I am kinda sure that promise will always resolve to true or false, so I can build a suite relying on it. If I do not provide a fail callback, then I'll get an Uncaught exception because of timeout, which will still fail my particular spec and still run the rest of the specs. It won't be uncaught by the way, Protractor is gonna catch it, but here I want to bring a second point, that Protractor is considered a tool which you use to write and run your code, and if an exception is caught by Protractor, then this exception has left your code unhandled and your code has a leak. But ... at the same time I do not think that one should waste time to catch everything in tests: if there is no element on a page or click has failed, then a respective spec will obviously fail too, which is fine in most of the cases. Unless you want to use the result of failure to build some code on top of it like it is in my sample.
That is the great thing about promises you are going to get a response, either an response of data or an error message. That extended to a series of promises like Webdriver uses you are going to get an array of responses or a failure response of the first one that fails. How you handle the failed response is up to you I usually just dump it into a log for the console to see what failed. The only thing you need to figure out is do you abort the rest of your tests or do you continue on.
Here is a decent article on the subject as well. http://www.toolsqa.com/selenium-webdriver/exception-handling-selenium-webdriver/
Fyi what you are doing is fine you are just never bothering to catch any of the errors though, I am not sure if that matters to you or not, you could also abstract the call in a function to auto handle the errors for you if you wanted to log them somewhere.

What does "Promise fires on the same turn of the event loop" mean?

New to NodeJS. Going through the promise tutorial ('promise-it-wont-hurt') I have the following script:
var Q = require('q');
var deferred = Q.defer();
deffered.resolve('SECOND');
deffered.promise.then(console.log);
console.log('FIRST');
The output:
FIRST
SECOND
I don't get it, I would have thought that since resolved is fired first I should see second first.
They explain that this happens because of 'Promise fires on the same turn of the event loop'. I don't understand what that means...
Basically, the point is that the promise's then handler will not run before the current flow of code has finished executing and control is returned to the execution environment (in this case Node).
This is an important characteristic of Promises/A+ compliant promises because it ensures predictability. Regardless of whether the promise resolves immediately:
function getAPromise() {
var Q = require('q');
var deferred = Q.defer();
deferred.resolve('SECOND');
return deferred.promise;
}
getAPromise().then(console.log);
console.log('FIRST');
or whether it resolves after 10 seconds:
function getAPromise() {
var Q = require('q');
var deferred = Q.defer();
setTimeout(function () {
deferred.resolve('SECOND');
}, 10000);
return deferred.promise;
}
getAPromise().then(console.log);
console.log('FIRST');
you can be assured that FIRST will always be logged first.
This is discussed at length in chapters 2 and 3 of You don't know JS - async & performance. Asynchronous operations that sometimes run asynchronously and sometimes run synchronously are called Zalgos, and they are not considered to be a good thing.
It's important to note that the widely used promises in jQuery do not obey this behavior and have a number of other problems as well. If you find yourself with a jQuery promise, wrap it in a proper promise and proceed:
Q($.ajax(...)).then(breatheWithEase);
They explain that this happens because of 'Promise fires on the same turn of the event loop'. I don't understand what that means...
Me neither, imho this part doesn't make much sense. It probably should mean that when you call resolve or reject, the promise is settled immediately, and from then on will not change its state. This has nothing to do with the callbacks yet.
What is more important to understand is the next sentence in that paragraph:
You can expect that the functions passed to the "then" method of a promise will be called on the NEXT turn of the event loop.
It's like a simple rule of thumb: then callbacks are always called asynchronously. Always, in every proper promise implementation; this is mandated by the Promises/A+ specification.
Why is this? For consistency. Unlike in your example snippet, you don't know how or when a promise is resolved - you just get handed it back from some method you called. And it might have already been resolved, or it might be still pending. Now you call the promise's then method, and you can know: it will be asynchronous. You don't need to deal with cases that might be synchronous or not and change the meaning of your code, it simply is always asynchronous.

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