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);